Matematik T4

Matematik Tingkatan 4 Bab 6 79 (c) Sebuah bakul buah-buahan mesti diisi dengan oren dan lemon. Jumlah bilangan oren dan lemon tidak boleh lebih daripada 20 biji dan bilangan oren mesti lebih daripada bilangan lemon sekurang-kurangnya 2 biji. A fruit basket must be filled with oranges and lemons. The total number of oranges and lemons must not exceed 20 and the number of oranges must be at least more than the number of lemon by 2. (i) Tuliskan semua ketaksamaan linear selain daripada x M 0 dan y M 0 untuk mewakili cara mengisi bakul buah-buahan. Write all the linear inequalities other than x M 0 and y M 0 to represent the way of filling up the fruit basket. Katakan bilangan oren = x dan bilangan lemon = y. Let the number of oranges = x and the number of lemons = y. x + y N 20 x – y M 2 y N 20 – x y N x – 2 (d) Sebuah pasukan badminton akan dihantar ke daerah lain untuk latihan lanjutan. Bilangan maksimum atlet dalam pasukan itu ialah 10 orang dan bilangan lelaki mesti kurang daripada bilangan perempuan sebanyak 2 orang. A badminton team will be sent to another district for advance training. The maximum number of athletes in the team is 10 and the number of men must be less than the number of women by 2. (i) Tuliskan semua ketaksamaan linear selain x M 0 dan y M 0 untuk mewakili pembentukan pasukan badminton itu. Write all the linear inequalities besides x M 0 and y M 0 to represent the formation of the badminton team. Katakan x = bilangan lelaki dan y = bilangan perempuan. Let x = the number of men and y = the number of women. x + y N 10 y – x > 2 y N 10 – x y > x + 2 (ii) Lukis dan lorekkan rantau yang memuaskan sistem ketaksamaan linear ini. Draw and shade the region which satisfies the system of linear inequalities. y N 20 – x y N x – 2 x 0 20 y 20 0 x 2 4 y 0 2 x y 20 O 2 20 y = 20 – x y = x – 2 (iii) Daripada graf, tentukan sama ada 10 biji lemon dapat diisikan dalam bakul buahbuahan tersebut dengan syarat yang ditetapkan. Berikan justifikasi anda. From the graph, determine whether 10 lemons can be filled into the fruit basket with the specified conditions. Give your justification. Tidak, kerana y = 10 berada di luar rantau berlorek. No, because y = 10 is outside of the shaded region. (ii) Lukis dan lorekkan rantau yang memuaskan sistem ketaksamaan linear ini. Draw and shade the region which satisfies the system of linear inequalities. y N 10 – x y > x + 2 x 0 10 y 10 0 x –2 0 2 y 0 2 4 x y 10 2 O 10 y = 10 – x y = x + 2 (iii) Daripada graf, tentukan bilangan maksimum lelaki yang boleh menyertai pasukan badminton itu. From the graph, determine the maximum number of men who can join the badminton team. Bilangan maksimum lelaki yang boleh menyertai pasukan badminton ialah 4 orang. The maximum number of men who can join the badminton team is 4. CONTOH

Matematik Tingkatan 4 Bab 6 80 Kertas 1 Jawab semua soalan. / Answer all questions. 1 Rajah 1 menunjukkan satu rantau yang memuaskan ketaksamaan linear P. Diagram 1 shows a region which satisfies linear inequality P. O 7 7 x y Rajah 1/ Diagram 1 Tentukan ketaksamaan linear P pada Rajah 1. Determine linear inequality P in Diagram 1. A x + y N 7 B x + y M 7 C x – y < 7 D x – y > 7 2 Rajah 2 menunjukkan satu sistem ketaksamaan linear pada suatu satah Cartes. Diagram 2 shows a system of linear inequalities on a Cartesian plane. x + 2y = 10 x y 10 5 O 3y = x Rajah 2/ Diagram 2 Tentukan ketaksamaan linear yang memuaskan rantau berlorek pada satah Cartes. Determine the linear inequalities which satisfy the shaded region on the Cartesian plane. A 3y > x dan/ and x + 2y M 10 B 3y > x dan/ and x + 2y > 10 C 3y N x dan/ and x + 2y M 10 D 3y M x dan/ and x + 2y N 10 3 Rajah 3 menunjukkan dua ketaksamaan linear dilukis pada suatu satah Cartes. Diagram 3 shows two linear inequalities drawn on a Cartesian plane. O 1 12 6 –3 x y A D C B Rajah 3/ Diagram 3 Antara rantau A, B, C dan D, yang manakah memuaskan ketaksamaan linear 3x – y M 3 dan x + 2y M 12? Which region A, B, C and D satisfies the linear inequalities 3x – y M 3 and x + 2y M 12? 4 Rajah 4 menunjukkan satu sistem ketaksamaan linear pada suatu satah Cartes. Diagram 4 shows a system of linear inequalities on a Cartesian plane. y R Q P S x O –2 2 4 Rajah 4/ Diagram 4 Antara rantau berikut, yang manakah memuaskan ketaksamaan linear y M x – 2 dan 2x + y N 4? Which of the following regions satisfies the linear inequalities y M x – 2 and 2x + y N 4? A P C R B Q D S Praktis Kendiri CONTOH

Matematik Tingkatan 4 Bab 6 81 5 Rajah 5 menunjukkan satu sistem ketaksamaan linear pada suatu satah Cartes. Diagram 5 shows a system of linear inequalities on a Cartesian plane. y x –2 O 5 5 1 Rajah 5/ Diagram 5 Antara berikut, yang manakah memuaskan rantau berlorek pada satah Cartes? Which of the following satisfy the shaded region on the Cartesian plane? A 2y > x + 2, y < 5 dan/ and x + y < 5 B 2y < x + 2, y < 5 dan/ and x + y N 5 C 2y N x + 2, y < 5 dan/ and x + y N 5 D 2y M x + 2, y < 5 dan/ and x + y < 5 6 Rajah 6 menunjukkan satu sistem ketaksamaan linear yang diwakili pada suatu satah Cartes. Diagram 6 shows a system of linear inequalities represented on a Cartesian plane. 2x + y = 3 y – x = 4 O 5 x y Rajah 6/ Diagram 6 Antara ketaksamaan linear berikut, yang manakah tidak memuaskan rantau berlorek pada satah Cartes? Which of the following linear inequalities does not satisfy shaded region on the Cartesian plane? A x < 5 B y < 5 C y – x N 4 D 2x + y > 3 Kertas 2 Jawab semua soalan. / Answer all questions. Bahagian A/ Section A 1 (a) Pada graf, lorek rantau yang memuaskan kedua-dua ketaksamaan linear 4x + 5y N 20 dan 3y > x. On the graph, shade the region which satisfies the two linear inequalities 4x + 5y N 20 and 3y > x. [2 markah/ marks] x y 3y = x 4x + 5y = 20 O (b) Sebuah beg mempunyai 25 biji guli yang terdiri daripada dua warna berlainan. Bilangan guli merah adalah 2 kali lebih daripada bilangan guli hijau. Bilangan minimum guli hijau ialah 5 biji. Tuliskan tiga ketaksamaan linear selain x M 0 dan y M 0 bagi sistem ketaksamaan linear dalam situasi ini. A bag consists of 25 marbles which are made up of two different colours. The number of red marbles are 2 times more than the number of green marbles. The minimum number of green marbles is 5. Write three linear inequalities other than x M 0 and y M 0 for the system of linear inequalities in this situation. [3 markah/ marks] Katakan x = bilangan guli merah dan y = bilangan guli hijau. Let x = the number of red marbles and y = the number of green marbles. x + y N 25, x > 2y, y M 5 CONTOH

Matematik Tingkatan 4 Bab 6 82 2 Pada graf, lorek rantau yang memuaskan ketiga-tiga ketaksamaan linear y N –x, y N x – 2 dan 4x – 9y < 36. On the graph, shade the region which satisfies all three linear inequalities y N –x, y N x – 2 and 4x – 9y < 36. [3 markah/ marks] x y O y = –x y = x – 2 4x – 9y = 36 3 Rajah 1 menunjukkan satu sistem ketaksamaan linear pada suatu satah Cartes. Diagram 1 shows a system of linear inequalities on a Cartesian plane. 3 (6, 3) (–6, –3) –3 y x O 1 2 3 Rajah 1/ Diagram 1 (a) Tuliskan tiga ketaksamaan linear dalam sistem ketaksamaan linear itu. Write three linear inequalities in the system of linear inequalities. [3 markah/ marks] Persamaan / Equation : Kecerunan/ Gradient, m: 0 ∴ y < 3 Persamaan ➁/ Equation ➁: Kecerunan/ Gradient, m: – 3 –3 = 1 ∴ y N x + 3 Persamaan ➂/ Equation ➂: Kecerunan/ Gradient, m: 3 – (–3) 6 – (–6) = 1 2 ∴ y M 1 2 x (b) Dari graf, nyatakan julat bagi x jika y = 2. From the graph, state the range of x if y = 2. [1 markah/ mark] –1 N x N 4 CONTOH

Matematik Tingkatan 4 Bab 6 83 Bahagian B/ Section B 4 Sebuah kilang kertas menghasilkan dua jenis kertas, iaitu kertas 70 gram dan kertas 80 gram. Semua kertas yang dihasilkan dibungkus di dalam sebuah kotak. Masa operasi penghasilan kertas kilang tersebut adalah 8 jam sehari. Syarat-syarat penghasilan kertas adalah seperti berikut: A paper factory produces two types of papers, which are 70 grams papers and 80 grams papers. All the papers produced are packed into a box. The operation time of paper production of the factory is 8 hours a day. The conditions of paper production are as follows: I Mesin mengambil masa 5 minit untuk menghasilkan sekotak kertas 70 gram manakala mesin mengambil masa 8 minit untuk menghasilkan sekotak kertas 80 gram. The machine takes 5 minutes to produce a box of 70 grams papers while the machine takes 8 minutes to produce a box of 80 grams papers. II Bilangan kotak kertas 80 gram yang terhasil dalam sehari tidak kurang daripada bilangan kotak kertas 70 gram. The number of boxes of 80 grams papers produced in a day is not less than the number of boxes of 70 grams papers. III Bilangan minimum kertas 80 gram yang terhasil dalam sehari adalah lebih daripada 30 kotak. The minimum number of 80 grams papers produced in a day is more than 30 boxes. (a) Tuliskan tiga ketaksamaan linear selain x M 0 dan y M 0 yang mewakili penghasilan kertas kilang tersebut dalam sehari. Write three linear inequalities other than x M 0 and y M 0 which represent the paper production of the factory in one day. [3 markah/ marks] Katakan x = bilangan kotak kertas 70 gram dan y = bilangan kotak kertas 80 gram. Let x = the number of boxes of 70 grams papers and y = the number of boxes of 80 grams papers. 5x + 8y N 480 y M x y M 30 8y N –5x + 480 y N – 5 8 x + 60 (b) Lukis dan lorekkan rantau sepunya yang memuaskan tiga ketaksamaan linear pada grid segi empat sama di bawah. Draw and shade the common region which satisfy the three linear inequalities on the square grids below. [4 markah/ marks] y N – 5 8 x + 60 x y 60 30 O 96 y = x y = 30 y = – 5 8x + 60 (c) Dari graf, tentukan sama ada syarat penghasilan boleh dipatuhi sekiranya 80 kotak kertas 70 gram perlu dihasilkan. Beri justifikasi anda. From the graph, determine whether the conditions of production can be complied if 80 boxes of 70 grams papers need to be produced. Give your justification. [2 markah/ marks] Tidak, kerana nilai x = 80 berada di luar rantau berlorek. No, because the value of x = 80 is outside of the shaded region. x 0 80 96 y 60 10 0 CONTOH

84 Praktis Intensif 7.1 Graf Jarak-Masa Distance-Time Graphs Buku Teks m/s 184 – 194 1 Lukis graf jarak-masa berdasarkan jadual berikut. SP: 7.1.1 TP1 Mudah Draw a distance-time graph based on the table below. (a) Masa (saat) Time (second) 0 5 10 15 20 Jarak (m) Distance (m) 6 8 10 12 14 Jarak (m) Distance (m) Masa (saat) Time (second) 4 5 10 15 20 O 8 12 16 (b) Masa (jam) Time (hour) 0 1 2 3 4 Jarak (km) Distance (km) 50 100 150 200 250 Jarak (km) Distance (km) Masa (jam) Time (hour) 100 1 2 3 4 O 200 300 400 (c) Masa (minit) Time (minute) 0 15 30 45 60 Jarak (km) Distance (km) 0 25 50 75 100 Jarak (km) Distance (km) Masa (minit) Time (minute) 25 15 30 45 60 O 50 75 100 Masa (jam) Time (hour) 0 0.5 1.0 1.5 2.0 Jarak (km) Distance (km) 0 10 20 30 40 Jarak (km) Distance (km) Masa (jam) Time (hour) 10 0.5 1.0 1.5 2.0 O 20 30 40 Contoh Info Digital 7.1 7 Bab Graf Gerakan Graphs of Motion CONTOH

Matematik Tingkatan 4 Bab 7 85 2 Lukis graf jarak-masa berdasarkan maklumat berikut. SP: 7.1.1 TP2 TP3 Mudah Draw a distance-time graph based on the following information. (a) Encik Ramli bertolak dari Kuala Lumpur untuk menghadiri konvokesyen anak perempuannya di Melaka. Beliau mengambil masa 3 jam untuk memandu sejauh 150 km. Diberi s = 50t, lukis graf jarak-masa. Encik Ramli travels from Kuala Lumpur to Melaka to attend his daughter’s convocation. He takes 3 hours to travel 150 km. Given that s = 50t, draw a distance-time graph. Masa, t (jam) Time, t (hour) 0 1 2 3 Jarak, s (km) Distance, s (km) 0 50 100 150 Jarak (km) Distance (km) Masa (jam) Time (hour) 50 1 2 3 O 100 150 (b) Suatu zarah bergerak dari titik P ke titik Q sejauh 25 mm. Persamaan bagi jarak zarah dari titik P ke titik Q untuk tempoh 5 saat ialah s = 5t, dengan keadaan s ialah jarak dalam mm dan t ialah masa dalam saat. A particle moves from point P to point Q with a distance of 25 mm. The equation of the distance of the particle from point P to point Q for 5 seconds is s = 5t, where s is the distance in mm and t is the time in second. Masa, t (saat) Time, t (second) 0 1 2 3 4 5 Jarak, s (mm) Distance, s (mm) 0 5 10 15 20 25 Jarak (mm) Distance (mm) Masa (saat) Time (second) 10 1 2 3 4 5 O 20 30 Benjamin telah menyertai satu maraton. Jumlah jarak maraton tersebut ialah 20 km. Diberi larian Benjamin mengikut s = 20 – 1 9 t dengan keadaan s ialah jarak dalam km dan t ialah masa dalam minit. Lukis graf jarak-masa bagi 0 N t N 180 min. Benjamin participated in a marathon. The total distance of the marathon is 20 km. Given that Benjamin's running follows s = 20 − 1 9 t, where s is the distance in km and t is the time taken in minute. Draw a distance-time graph for 0 N t N 180 min. Masa, t (minit) Time, t (minute) 0 180 Jarak, s (km) Distance, s (km) 20 0 Jarak (km) Distance (km) Masa (minit) Time (minute) 5 30 60 180 O 10 15 20 Tip Bestari Suatu garis lurus boleh dilukis dengan memplot sekurangkurangnya dua titik yang berkaitan. A straight line can be drawn by plotting at least two related points. Contoh CONTOH

Matematik Tingkatan 4 Bab 7 86 3 Tafsirkan graf jarak-masa dengan menyelesaikan setiap masalah yang berikut. SP: 7.1.2 TP4 Sukar Interpret the distance-time graph by solving each of the following problems. Graf jarak-masa menunjukkan perjalanan Arif dari bandar M ke bandar N. The distance-time graph shows Arif’s journey from town M to town N. Arif berhenti sebentar di bandar N sebelum pulang ke bandar M. Arif stops for a while at town N before going back to town M. (a) Nyatakan jarak, dalam km, dari bandar M ke bandar N. State the distance, in km, from town M to town N. (b) Berapa lamakah Arif berhenti di bandar N? How long does Arif stop at town N? (c) Hitung laju kereta Arif pada 6 minit yang pertama, dalam km j–1. Calculate the speed of Arif’s car in the first 6 minutes, in km h–1. (d) Huraikan gerakan kereta dalam tempoh 7 minit terakhir. Describe the motion of the car in the last 7 minutes. (a) 14 – 0 = 14 km (b) 13 – 6 = 7 minit/ minutes (c) Laju = Jarak Masa Speed = Distance Time = 14 6 60 1 minit = 1 60 jam 1 minute = 1 60 hour = 140 km j–1/ km h–1 Jarak (km) Distance (km) Masa (minit) Time (minute) 14 6 13 20 O (a) Graf jarak-masa menunjukkan perjalanan sebuah bas dalam tempoh 45 minit. The distance-time graph shows the journey of a bus in 45 minutes. (i) Hitung/ Calculate (a) tempoh masa bas itu berhenti. the period of time that the bus is stationary. (b) laju, dalam km j–1, pada 5 minit yang pertama. the speed, in km h–1, in the first 5 minutes. (c) nilai s, jika laju purata bas tersebut ialah 60 km j–1. the value of s, if the average speed of the bus is 60 km h–1. (ii) Huraikan gerakan bas itu dalam tempoh 20 minit terakhir. Describe the motion of the bus in the last 20 minutes. (i) (a) 25 – 5 = 20 minit/ minutes (b) Laju/ Speed = 7 – 4 5 – 0 60 = 36 km j–1 /km h–1 (c) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time 60 = s 45 60 s = 60 × 45 60 = 45 km Jarak (km) Distance (km) Masa (minit) Time (minute) 4 7 s 5 25 45 O (d) Laju/ Speed = Jarak/ Distance Masa/ Time = 14 20–13 60 7 minit terakhir Last 7 minutes = 120 km j–1/ km h–1 Kereta tersebut bergerak dengan laju 120 km j–1 dalam tempoh 7 minit terakhir. The car move with a speed of 120 km h–1 in the last 7 minutes. (ii) Laju/ Speed = 45 – 7 45 – 25 60 = 114 km j–1 /km h–1 Bas itu bergerak sejauh 38 km dengan kelajuan 114 km j–1 dalam tempoh 20 minit terakhir. The bus travelled 38 km with a speed of 114 km h–1 in the last 20 minutes. Contoh Tip Bestari km j–1 juga boleh ditulis sebagai km/j km h–1 also can be written as km/h CONTOH

Matematik Tingkatan 4 Bab 7 87 (b) Graf jarak-masa berikut menunjukkan pergerakan zarah A dan zarah B dalam tempoh 25 saat. The following distance-time graphs show the movement of particle A and particle B in 25 seconds respectively. Jarak (m) Distance (m) Masa (s) 25 Time (s) 36 186 A B O 25 O Jarak (m) Distance (m) Masa (s) Time (s) 225 Hitung beza laju, dalam m s–1, kedua-dua zarah itu. Calculate the difference of speed, in m s–1, of both particles. Laju zarah A/ The speed of particle A: 186 – 36 25 = 6 m s–1 Laju zarah B/ The speed of particle B: 225 25 = 9 m s–1 Beza laju/ Difference of speed: 9 – 6 = 3 m s–1 (c) Graf jarak-masa menunjukkan perjalanan sebuah bas dari stesen X ke stesen Y dalam tempoh 80 minit. The distance-time graph shows the journey of a bus from station X to station Y for a period of 80 minutes. Jarak (km) Distance (km) Masa (minit) 20 Time (minute) 35 60 100 O 50 80 (i) Huraikan gerakan bas itu dalam 20 minit yang pertama. Describe the movement of the bus in the first 20 minutes. (ii) Hitung laju, dalam km j–1, bagi 30 minit terakhir. Calculate the speed, in km h–1, for the last 30 minutes. (i) Laju/ Speed = 60 – 35 20 60 = 75 km j–1 / km h–1 Bas itu bergerak sejauh 25 km dalam tempoh 20 minit dengan kelajuan 75 km j–1. The bus travelled 25 km in 20 minutes with a speed of 75 km h–1. (ii) Laju/ Speed = 100 – 60 80 – 50 60 = 80 km j–1/ km h–1 (d) Graf jarak-masa berikut mewakili perjalanan Dedrick dari Georgetown ke Batu Ferringhi dan perjalanan David dari Batu Ferringhi ke Georgetown. Diberi masa mereka bertolak adalah sama. The following distance-time graph represents Dedrick’s journey from Georgetown to Batu Ferringhi and David’s journey from Batu Ferringhi to Georgetown. Given that they departed at the same time. Jarak (km) Distance (km) Masa (minit) Time (minute) 5 6 12 O 20 25 30 45 David Dedrick (i) Hitung tempoh masa, dalam minit, kereta David berhenti. Calculate the length of time, in minutes, David's car stopped. (ii) Jika perjalanan mereka bermula pada pukul 9 a.m., pada pukul berapakah Dedrick akan bertemu dengan David? If their journey starts at 9 a.m., what is the time Dedrick meets David? (iii) Hitung laju, dalam km j–1, Dedrick apabila dia bertemu dengan David. Calculate the speed, in km h–1, of Dedrick when he meets David. (i) 30 – 25 = 5 minit/ minutes (ii) 9.20 a.m. (iii) Laju/ Speed = 6 20 60 = 18 km j–1/ km h–1 CONTOH

Matematik Tingkatan 4 Bab 7 88 4 Selesaikan masalah berikut. SP: 7.1.3 TP4 Sukar Solve the following problems. Muthu berjalan kaki ke sekolah pada setiap pagi. Graf jarak-masa berikut menunjukkan perjalanannya. Muthu akan bertemu dengan Ahmad di rumah Ahmad sebelum mereka bertolak ke sekolah bersama-sama. Muthu walks to school every morning. The following distancetime graph shows his journey. Muthu will meet Ahmad at Ahmad’s house before they go to school together. Masa (sistem 24 jam) Time (24 hour system) 0.28 0.56 0714 0719 0728 Jarak (km) Distance (km) 0700 O (a) Hitung laju, dalam km j–1, Muthu dari rumahnya ke rumah Ahmad. Calculate the speed, in km h–1, of Muthu from his house to Ahmad’s house. (b) Nyatakan masa, dalam minit, Muthu menunggu Ahmad di rumah Ahmad. State the duration of time, in minutes, Muthu waits for Ahmad at Ahmad’s house. (c) Jika mereka sampai di sekolah pada 7.28 a.m., hitung laju purata Muthu, dalam km j–1. If they reach school at 7.28 a.m., calculate Muthu’s average speed, in km h–1. (a) Masa/ Time = 0714 – 0700 = 14 minit/ minutes Laju/ Speed = 0.28 14 60 = 1.2 km j–1/ km h–1 (b) Masa/ Time = 0719 – 0714 = 5 minit/ minutes (c) Laju purata = Jumlah jarak dilalui Jumlah masa diambil Average speed = Total distance travelled Total time taken = 0.56 28 60 = 1.2 km j–1/ km h–1 Encik Lim memandu dari rumahnya ke bandar X sebelum meneruskan perjalanannya ke bandar Y. Graf jarak-masa berikut menunjukkan perjalanan Encik Lim. Mr Lim droves from his house to town X before continues his driving to town Y. The following distance-time graph shows Mr Lim’s journey. Jarak (km) Distance (km) Masa (jam) Time (hour) Bandar X Town X Bandar Y Town Y O s 20 0.5 2.5 4.5 180 (a) Hitung laju purata, dalam km j–1, bagi perjalanan Encik Lim dari rumahnya ke bandar Y. Calculate the average speed, in km h–1, for Mr Lim’s journey from his house to town Y. (b) Jika kadar perubahan jarak terhadap masa pemanduan Encik Lim dari rumahnya ke bandar X ialah 20 km j–1, berapakah jarak, dalam km, di antara bandar X dengan bandar Y? If the rate of change of distance to time of Mr Lim's driving from his house to town X is 20 km h–1, what is the distance, in km, between town X and town Y? (c) Huraikan gerakan kereta dari rumah Encik Lim ke bandar Y. Describe the movement of the car from Mr Lim’s house to town Y. (a) Laju purata = Jumlah jarak dilalui Jumlah masa diambil Average speed = Total distance travelled Total time taken = 180 km 4.5 j/ h = 40 km j–1/ km h–1 (b) Jumlah jarak = Laju purata × Jumlah masa Total distance = Average speed × Total time = 20 × 2.5 = 50 km Jarak di antara bandar X dengan bandar Y: The distance between town X and town Y: 180 – 50 = 130 km (c) Kereta bergerak sejauh 180 km dalam tempoh 4.5 jam dengan laju purata 40 km j–1. The car travelled 180 km in 4.5 hours with an average speed of 40 km h–1. Contoh CONTOH

Matematik Tingkatan 4 Bab 7 89 7.2 Graf Laju-Masa Speed-Time Graphs Buku Teks m/s 195 – 205 1 Lukis graf laju-masa berdasarkan maklumat yang diberi. SP: 7.2.1 TP1 Mudah Draw the speed-time graph based on the given information. (a) Masa (saat) Time (second) 0 2 4 6 8 Laju (m s–1) Speed (m s–1) 8 6 4 2 0 Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 2 2 4 6 8 O 4 6 8 (b) Masa (saat) Time (second) 0 10 20 30 40 Laju (m s–1) Speed (m s–1) 0 5 10 15 20 Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 5 10 20 30 40 O 10 15 20 (c) Masa (saat) Time (second) 0 3 6 9 12 Laju (m s–1) Speed (m s–1) 40 30 20 10 0 Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 10 3 6 9 12 O 20 30 40 Masa (saat) Time (second) 0 3 6 9 12 Laju (m s–1) Speed (m s–1) 0 5 10 15 20 Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 5 3 6 9 12 O 10 15 20 Contoh Info Digital 7.2 CONTOH

Matematik Tingkatan 4 Bab 7 90 2 Hitung jarak yang dilalui oleh suatu zarah bagi graf laju-masa berikut. SP: 7.2.2 TP2 TP3 Sederhana Calculate the distance travelled by a particle in the following speed-time graphs. (a) Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) O 9 12 8 10 Jarak/ Distance = [ 1 2 × (8 + 10) × 9] + [ 1 2 × 8 × (12 – 9)] = 81 + 12 = 93 m (b) Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) O 2 8 3 12 10 Jarak/ Distance = [ 1 2 × (3 + 12) × 2] + (6 × 12) + [ 1 2 × 2 × 12] = 15 + 72 + 12 = 99 m (c) Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) O 3 7 9 7 12 Jarak/ Distance = [ 1 2 × (7 + 4) × 7] + [ 1 2 × (7 + 12) × 2] = 38.5 + 19 = 57.5 m (d) 7 Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) O 6 10 12 Jarak/ Distance = [ 1 2 × (6 + 10) × 7] + [ 1 2 × 5 × 10] = 56 + 25 = 81 m Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) 15 21 O 7 Jarak = Luas trapezium Distance = The area of trapezium = 1 2 × (15 + 21) m s–1 × 7 s Unit sama, saat, s Same unit, second, s = 126 m Contoh Tip Bestari ·Pastikan unit yang digunakan untuk masa pada laju dan masa adalah sama Make sure that the unit used for time in speed and time is the same ·Luas di bawah graf = Jarak yang dilalui The area under a graph = The distance travelled CONTOH

Matematik Tingkatan 4 Bab 7 91 3 Selesaikan masalah berikut. SP: 7.2.2 TP4 Sukar Solve the following problems. Graf laju-masa berikut menunjukkan kelajuan Kamal semasa berjoging. Jumlah jarak yang dilalui oleh Kamal semasa joging ialah 298 m. The following speed-time graph shows Kamal’s speed during his jog. The total distance of Kamal’s jog is 298 m. Masa (saat) Time (seconds) Laju (m s–1) Speed (m s–1) 15 23 t 18 O Hitung/ Calculate (a) nilai t. the value of t. (b) laju purata, dalam m s–1, zarah itu dalam masa 18 saat. the average speed, in m s–1, of the particle in 18 seconds. (a) [ 1 2 × (15 + 23) × t] + [(18 – t) × 15] = 298 19t + 270 – 15t = 298 19t – 15t = 298 – 270 4t = 28 t = 7 (b) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time = 298 18 = 16.56 m s–1 Graf laju-masa berikut menunjukkan pergerakan suatu zarah dalam 10 saat. The following speed-time graph shows the movement of a particle in 10 seconds. Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) 36 33 18 O 4 10 Hitung/ Calculate (a) jumlah jarak yang dilalui oleh zarah itu dalam tempoh 10 saat. the total distance travelled by the particle in 10 seconds. (b) laju purata, dalam m s–1, zarah itu dalam masa 10 saat. the average speed, in m s–1, of the particle in 10 seconds. (a) Jumlah jarak/ Total distance = Luas di bawah graf/ The area under the graph = [ 1 2 × (18 + 36) m s–1 × 4 s] + [ 1 2 × (18 + 33) m s–1 × (10 – 4) s] = 108 + 153 = 261 m (b) Laju purata = Jumlah jarak dilalui Jumlah masa diambil Average speed = Total distance travelled Total time taken = 261 10 = 26.1 m s–1 Contoh CONTOH

Matematik Tingkatan 4 Bab 7 92 4 Selesaikan setiap masalah yang berikut dengan mentafsirkan graf laju-masa. SP: 7.2.3 TP4 Sukar Solve each of the following problems by interpreting the speed-time graph. Danial menunggang motosikal ke kedai runcit. Graf laju-masa berikut menunjukkan gerakan Danial dari rumah ke persimpangan lampu isyarat sebelum sampai ke kedai runcit. Danial rides a motorcycle to the grocery shop. The following speed-time graph shows Danial’s movement from his house to a traffic light junction before he arrives at the grocery shop. Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) 15 12 15 5 5 O (a) Huraikan gerakan motosikal Danial bagi tempoh 5 saat pertama. Describe Danial’s motorcycle movement for the period of the first 5 seconds. (b) Apakah yang berlaku terhadap gerakan motosikal Danial dari saat ke-5 hingga saat ke-12? What happens to Danial’s motorcycle movement from 5th second to 12th second? (c) Hitung kadar perubahan laju terhadap masa, dalam m s–2, bagi 3 saat terakhir. Calculate the rate of change of speed to time, in m s–2, for the last 3 seconds. (d) Hitung jarak, dalam m, yang dilalui semasa nyahpecutan dan huraikan gerakan motosikal pada tempoh tersebut. Calculate the distance, in m, travelled during deceleration and describe the motorcycle movement in the period. (a) Kadar perubahan laju untuk tempoh 5 saat pertama The rate of change of speed for the first 5 seconds = Perubahan laju Perubahan masa = Change of speed Change of time = 15 – 5 5 – 0 = 2 m s–2 Motosikal memecut dengan kadar 2 m s–2 dalam tempoh 5 saat pertama. The motorcycle accelerates at a rate of 2 m s–2 in the first 5 seconds. (b) Motosikal Danial bergerak dengan laju seragam 15 m s–1 dari saat ke-5 hingga saat ke-12. Danial’s motorcycle moves at a uniform speed of 15 m s–1 from the 5th second to the 12th second. (c) Kadar perubahan laju untuk tempoh 3 saat terakhir The rate of change of speed for the last 3 seconds = Perubahan laju Perubahan masa = Change of speed Change of time = 0 – 15 15 – 12 = –5 m s–2 (d) Jarak yang dilalui = Jarak dalam 3 saat terakhir Distance travelled = Distance in the last 3 seconds = 1 2 × 3 × 15 = 22.5 m Motosikal bergerak sejauh 22.5 m dalam tempoh 3 saat dengan nyahpecutan 5 m s–2. The motorcycle travelled 22.5 m in 3 seconds with a deceleration of 5 m s–2. Contoh Boleh ditulis sebagai/ Can be written as • pecutan/ acceleration = –5 m s–2 atau/ or • nyahpecutan/ deceleration = 5 m s–2 CONTOH

Matematik Tingkatan 4 Bab 7 93 (a) Graf laju-masa menunjukkan gerakan sebuah kereta dari bandar X ke bandar Y. The speed-time graph shows the movement of a car from town X to town Y. (i) Hitung kadar perubahan laju terhadap masa, dalam m min–2, bagi 40 saat yang pertama. Calculate the rate of change of speed to time, in m min–2, for the first 40 seconds. (ii) Huraikan gerakan kereta itu dari saat ke-40 hingga saat ke-80. Describe the movement of the car from the 40th second to the 80th second. (iii) Hitung jumlah jarak, dalam m, yang dilalui dalam 120 saat. Calculate the total distance, in m, travelled in 120 seconds. (i) Kadar perubahan laju untuk tempoh 40 saat pertama The rate of change of speed for the first 40 seconds = Perubahan laju Perubahan masa = Change of speed Change of time = 70 – 0 40 – 0 60 = 105 m min–2 Laju (m min–1) Speed (m min–1) Masa (saat) Time (second) 80 70 110 40 120 O (ii) Kereta bergerak dengan laju seragam 70 m min–1 dari saat ke-40 hingga saat ke-80. The car moves at a uniform speed of 70 m min–1 from the 40th second to the 80th second. (iii) Jumlah jarak/ Total distance: [ 1 2 × ( 40 + 80 60 ) × 70] + [ 1 2 ×(70 + 110 ) × 40 60 ] = 70 + 60 = 130 m (b) Graf jarak-masa berikut menunjukkan pergerakan sebuah kereta dari bandar Q ke bandar P. The following speed-time graph shows the movement of a car from town Q to town P. Laju (m min–1) Speed (m min–1) Masa (saat) Time (second) 70 90 7 21 35 O 35 (i) Hitung kadar perubahan laju terhadap masa, dalam m min–2, bagi tempoh 35 saat pertama. Calculate the rate of change of speed to time, in m min–2, for the first 35 seconds. (ii) Huraikan gerakan kereta dari saat ke-35 hingga saat ke-70. Describe the movement of the car from the 35th second to the 70th second. (iii) Hitung jumlah jarak, dalam m, yang dilalui dalam 90 saat. Calculate the total distance, in m, travelled in 90 seconds. (iii) Jumlah jarak/ Total distance: [ 1 2 (7 + 21) ( 35 60 )] + [21 (70 – 35 60 )] + [ 1 2 (21 + 35)(90 – 70 60 )] = 8.167 + 12.25 + 9.333 = 29.75 m (i) Kadar perubahan laju/ Rate of change of speed: 21 – 7 35 – 0 60 = 24 m min–2 (ii) Kereta bergerak dengan laju seragam 21 m min–1 dari saat ke-35 hingga saat ke-70. The car moves with a uniform speed of 21 m min–1 from the 35th second to the 70th second. CONTOH

Matematik Tingkatan 4 Bab 7 94 5 Selesaikan setiap masalah yang berikut. SP: 7.2.4 TP4 KBAT Sukar Solve each of the following problems. (a) Rajah berikut menunjukkan graf laju-masa Rickman pulang ke rumah dari tempat kerjanya dalam masa 120 s. The following diagram shows the speed-time graph for Rickman returning home from work within 120 s. 40 t 120 Laju (m s–1) Speed (m s–1) Masa (saat)/ Time (second) 92 12 O Hitung/ Calculate (i) tempoh masa, apabila dia memandu dengan menggunakan laju seragam, dalam sebutan t. the duration of time, when he travels with constant velocity in terms of t. (ii) kadar perubahan laju, dalam m s–2, Rickman dalam tempoh masa 40 s pertama. the rate of change of speed, in m s–2, of Rickman in the first 40 s. (iii) nilai t jika jumlah jarak yang dilalui oleh Rickman ialah 2 590 m. the value of t, if the total distance travelled by Rickman is 2 590 m. (iv) laju purata, dalam m s–1 bagi seluruh perjalanan. the average speed, in m s–1, of the whole journey. (b) Rajah berikut menunjukkan graf laju-masa bagi sebuah van dalam masa t jam. The following diagram shows the speed-time graph for the journey of a van within t hours. Laju (km j–1) Speed (km h–1) Masa (jam) Time (hour) 30 0.3 0.9 t O Hitung/ Calculate (i) pecutan, dalam km j–2, van itu dalam tempoh 0.3 jam pertama. the acceleration, in km h–2, of the van in the first 0.3 hours. (ii) jarak yang dilalui dalam tempoh 0.9 jam pertama. the distance travelled in the first 0.9 hours. (iii) nilai t jika laju purata van dalam tempoh t jam ialah 20 km j–1. the value of t if the average speed of the van within t hours is 20 km h–1. (i) Pecutan = Laju Masa Acceleration = Speed Time = 30 0.3 = 100 km j–2/ km h–2 (ii) Jarak/ Distance = 1 2 × (0.6 + 0.9) × 30 = 22.5 km (iii) Jumlah jarak/ Total distance: 22.5 + [ 1 2 × (t – 0.9) × 30] = 22.5 + 15t – 13.5 = 15t + 9 Laju purata/ Average speed = 20 km j–1/ km h–1 15t + 9 t = 20 15t + 9 = 20t 5t = 9 t = 1.8 (i) (t – 40) s (ii) Kadar perubahan laju/ Rate of change of speed: = Perubahan laju/ Change of speed Perubahan masa/ Change of time = 12 – 92 40 – 0 = –2 m s–2 (iii) [ 1 2 × (12 + 92) × 40] + [12(t – 40)] + [ 1 2 × 12 × (120 – t)] = 2 590 2 080 + (12t – 480) + (720 – 6t) = 2 590 2 320 + 6t = 2 590 6t = 270 t = 45 (iv) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time = 2 590 120 = 21.58 m s–1 CONTOH

Matematik Tingkatan 4 Bab 7 95 Kertas 1 Jawab semua soalan. / Answer all questions. 1 Rajah 1 menunjukkan graf jarak-masa bagi seorang penunggang motosikal yang bergerak dari bandar P ke bandar R. Diagram 1 shows the distance-time graph for a motorcyclist who travels from town P to town R. Jarak (km) Distance (km) Masa (jam) Time (hour) 0.3 0.7 0.9 9 6 O Rajah 1/ Diagram 1 Dia berhenti di bandar Q untuk berehat sebelum meneruskan perjalanannya ke bandar R. Nyatakan tempoh masa, dalam jam, semasa penunggang motor itu berhenti dalam perjalanannya. He stopped at town Q to rest before he continued his journey to town R. State the duration of time, in hours, when the motorcyclist stopped on his way. A 0.4 C 3.0 B 0.5 D 24 2 Rajah 2 menunjukkan sesaran suatu zarah. Diagram 2 shows the displacement of a particle. Masa/ Time (min) Jarak/ Distance (km) 60 30 20 45 90 110 O Rajah 2/ Diagram 2 Hitung laju purata, dalam km j–1, bagi zarah tersebut. Calculate the average speed, in km h–1, for the particle. A 52.01 B 49.09 C 38.18 D 32.07 Praktis Kendiri (c) Graf jarak-masa A menunjukkan pergerakan van A dalam tempoh masa 50 s manakala graf laju-masa B menunjukkan pergerakan van B dalam tempoh masa 50 s. The distance-time graph A shows the movement of van A in the period of 50 s whereas the speed-time graph B shows the movement of van B in the period of 50 s. Masa (saat) Time (second) 600 150 50 O Jarak (m) Distance (m) A: Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 10 5 25 50 O B: Cari/ Find (i) laju, dalam m s–1, van A dalam masa 50 s pertama. the speed, in m s–1, of van A in the first 50 s. (ii) laju purata, dalam m s–1, van B. the average speed, in m s–1, of van B. (iii) beza antara jarak, dalam m, yang dilalui oleh van A dan van B dalam tempoh masa 50 s. the difference between the distance, in m, travelled by van A and van B in the period of 50 s. (i) Laju/ Speed = 600 – 150 50 = 9 m s−1 (ii) Jarak/ Distance = [ 1 2 (5 + 10)(25)] + [(10)(25)] = 187.5 + 250 = 437.5 m Laju purata/ Average speed = 437.5 50 = 8.75 m s−1 (iii) Jarak A/ Distance A = 600 – 150 = 450 m Jarak B/ Distance B = 437.5 m Beza jarak/ Difference in distance: 450 – 437.5 = 12.5 m CONTOH

Matematik Tingkatan 4 Bab 7 96 3 Rajah 3 menunjukkan graf jarak-masa bagi dua buah kereta, A dan B. Diagram 3 shows the distance-time graph of two cars, A and B. Masa (minit) Time (minute) Jarak (m) Distance (m) A B 15 28 60 120 O 30 40 Rajah 3/ Diagram 3 Hitung beza laju purata, dalam m min–1, bagi keduadua kereta itu. Calculate the difference in the average speed, in m min–1, of both the cars. A – 51 9 C 9 7 B – 9 7 D 51 9 4 Rajah 4 menunjukkan graf laju-masa bagi suatu zarah. Diagram 4 shows the speed-time graph of a particle. Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) 3 9 12 12 O Rajah 4/ Diagram 4 Nyatakan tempoh masa, dalam saat, apabila zarah bergerak dengan laju seragam. State the duration of time, in second, when the particle moves with uniform speed. A 12 B 9 C 6 D 0.5 5 Rajah 5 menunjukkan graf laju-masa bagi sebuah kereta yang bergerak dari bandar A ke bandar B. Diagram 5 shows the speed-time graph for a car travelling from town A to town B. 10 20 25 30 O Laju (km min–1) Speed (km min–1) Masa (minit) Time (minute) Rajah 5/ Diagram 5 Cari kadar perubahan laju, dalam km min–2, bagi kereta itu dalam 10 minit terakhir. Find the rate of change of speed, in km min–2, of the car in the last 10 minutes. A 2.5 B 1.67 C –1.67 D –2.5 6 Rajah 6 menunjukkan graf laju-masa bagi suatu zarah yang bergerak dalam tempoh t saat. Diagram 6 shows the speed-time graph for a particle during a period of t seconds. Masa (saat) Time (second) Laju (m s–1) Speed (m s–1) 3 7 t 8 4 O Rajah 6/ Diagram 6 Hitung nilai t jika jumlah jarak yang dilalui ialah 62 m. Calculate the value of t if the total distance travelled is 62 m. A 3 B 4 C 10 D 12 CONTOH

Matematik Tingkatan 4 Bab 7 97 Kertas 2 Jawab semua soalan. / Answer all questions. Bahagian A/ Section A 1 Rajah 1 menunjukkan graf jarak-masa bagi gerakan sebuah kereta. Diagram 1 shows the distance-time graph of the movement of a car. Masa (saat) Time (second) Jarak (m) Distance (m) Z X Y 20 30 50 25 60 O Rajah 1/ Diagram 1 (a) Tentukan tempoh masa, dalam saat, ketika kereta itu berhenti. Determine the duration, in second, when the car is stopped. [1 markah/ mark] 30 – 20 = 10 s (b) Hitung laju, dalam m s–1, bagi kereta itu pada 20 s pertama. Calculate the speed, in m s–1, of the car in the first 20 s. [1 markah/ mark] Laju/ Speed = 25 20 = 1.25 m s–1 (c) Huraikan gerakan kereta itu bagi seluruh perjalanan. Describe the movement of the car for the whole journey. [2 markah/ marks] Laju purata/ Average speed = 60 50 = 1.2 m s–1 Kereta itu bergerak dengan laju purata 1.2 m s–1 dalam tempoh 50 saat sejauh 60 m. The car moved with an average speed of 1.2 m s–1 in 50 seconds for a distance of 60 m. 2 Jadual 1 menunjukkan perjalanan sebuah kereta. Table 1 shows the journey of a car. Masa Time Aktiviti Activity 0900 Memulakan perjalanan Start the journey 1015 Berhenti di R&R selepas memandu 80 km Stops at R&R after driving 80 km 1100 Meneruskan perjalanan sejauh 131 km lagi Continue the journey for another 131 km 1310 Tiba di destinasi Reach at the destination Jadual 1/ Table 1 (a) Lengkapkan rajah berikut. Complete the following diagram. [2 markah/ marks] Masa (minit) Time (minute) Jarak (km) Distance (km) 75 250 211 O h k (b) Cari nilai h dan k. Find the value of h and of k. [2 markah/ marks] h = 211 – 80 = 131 km k = 1100 – 0900 = 2 jam/ hours = 120 minit/ minutes (c) Hitung laju purata, in km j–1 , bagi perjalanan itu. Calculate the average speed, in km h–1, for the journey. [2 markah/ marks] Laju purata/ Average speed: 211 250 60 = 50.64 km j–1/ km h–1 CONTOH

Matematik Tingkatan 4 Bab 7 98 Bahagian B/ Section B 3 Rajah 2 menunjukkan sebuah graf jarak-masa bagi pergerakan kereta api A dan kereta api B antara bandar P dan bandar R. Diagram 2 shows a distance-time graph for the movement of train A and train B between city P and city R. Masa (sistem 24 jam)/ Time (24-hour system) Jarak (km) Distance (km) 3 7 B A 0730 0731 0732 0734 0735 0737 O Rajah 2/ Diagram 2 Kereta api A tiba di bandar Q selepas bertolak dari bandar P dalam masa 2 minit dan singgah di bandar Q selama 2 minit sebelum bertolak ke bandar R. Kereta api B bertolak dari bandar R ke bandar P, melalui bandar Q. Train A reached city Q after departing from city P in 2 minutes and stayed at city Q for another 2 minutes before departing to city R. Train B departed from city R to city P, passing through city Q. (a) Diberi kereta api B bergerak dengan laju v km j–1 dan tiba di bandar P dalam masa 3.5 minit. Hitung nilai v. Given that train B moved with a speed of v km h–1 and reached city P in 3.5 minutes. Calculate the value of v. [2 markah/ marks] v = (0 – 7) km 3.5 min = –7 km 3.5 min = –7 km (3.5 ÷ 60) j/ h = –120 km j–1/ km h–1 (b) Cari masa, dalam sistem 24 jam, kereta api B bertemu kereta api A di bandar Q. Find the time, in 24-hour system, that train B met train A in city Q. [4 markah/ marks] Jarak dari bandar R ke bandar Q: 7 – 3 = 4 km Distance from city R to city Q: Masa dari bandar R ke bandar Q: Time from city R to city Q: (4 ÷ 120 km j–1/ km h–1) × 60 min = 2 minit/ minutes Masa bertemu dengan kereta api A: The time met with train A: 0731 + 0002 = 0733 (c) Cari beza jarak, dalam km, yang dilalui antara kereta api A dan kereta api B apabila kedua-dua kereta api bertemu di bandar Q. Find the difference of distance, in km, travelled between train A and train B when both trains met in city Q. [1 markah/ mark] Jarak yang dilalui oleh kereta api A: 3 km Distance travelled by train A Jarak yang dilalui oleh kereta api B: Distance travelled by train B: 7 km – 3 km = 4 km Beza jarak/ Difference in distance: 4 – 3 = 1 km (d) Hitung laju purata, dalam km/j, bagi kereta api A. Calculate the average speed, in km/h, of train A. [2 markah/ marks] Laju purata/ Average speed: 7 km (0737 – 0730) ÷ 60 j/ h = 7 (7 ÷ 60) = 60 km/j / km/h 4 Dalam Rajah 3, OPQR mewakili perjalanan kereta A dan OSR mewakili perjalanan kereta B. Kedua-dua kereta bertolak dari bandar K pada masa yang sama. In Diagram 3, OPQR represents the journey for car A and OSR represents the journey for car B. Both cars left town K at the same time. Masa (jam) Time (hour) 100 30 0.4 1.2 1.6 2 P S Q R O Jarak (km) Distance (km) Rajah 3/ Diagram 3 (a) Nyatakan tempoh masa, dalam minit, kereta A dalam keadaan pegun. State the length of time, in minutes, that car A in stationary. [1 markah/ mark] 1.6 – 0.4 = 1.2 jam/ hours = 72 minit/ minutes CONTOH

Matematik Tingkatan 4 Bab 7 99 (b) Hitung laju purata, dalam km j–1, kereta A dalam masa 2 jam. Calculate the average speed, in km h–1, of car A in 2 hours. [2 markah/ marks] Laju purata/ Average speed: 100 2 = 50 km j–1/ km h–1 (c) Kedua-dua kereta bertolak pada jam 1200. Pada suatu masa, kedua-dua kereta bertemu. Both cars depart at 1200 hour. At a certain time, both cars meet. (i) Nyatakan masa kedua-dua kereta bertemu. State the time, when both cars meet. [2 markah/ marks] Di/ At S: 12 jam + 1.2 jam = 12 jam + 1 jam 12 minit 12 hours + 1.2 hours = 12 hours + 1 hour 12 minutes = 13 jam 12 minit = 13 hours 12 minutes = Jam 1312/ 1312 hour Di/ At R: 12 jam + 2 jam = 14 jam 12 hours + 2 hours = 14 hours = Jam 1400/ 1400 hour ∴ Kedua-dua kereta bertemu di S pada jam 1312 dan di R pada jam 1400. Both cars meet in S at 1312 hour and in R at 1400 hour. (ii) Nyatakan tempoh masa sebelum kedua-dua kereta bertemu pada kali kedua. State the length of time before both of the cars meet for the second time. [1 markah/ mark] 1400 – 1312 = 48 minit/ minutes (d) Lakarkan sebuah graf laju-masa bagi pergerakan kereta A dalam masa 2 jam. Sketch a speed-time graph for the movement of car A in 2 hours. [3 markah/ marks] Laju pada OP/ Speed at OP: 30 km 0.4 j/ h = 75 km/j/ km/h Laju pada QR/ Speed at QR: (100 – 30) km (2 – 1.6) j/ h = 70 0.4 = 175 km/j/ km/h Laju (km/j) Speed (km/h) Masa (jam) Time (hour) 175 75 0.4 1.6 2.0 O 5 Rajah 4 menunjukkan gerakan sebuah kereta dalam tempoh 13 saat. Diagram 4 shows the movement of a car within 13 seconds. Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 10 13 8 8 4 v t O Rajah 4/ Diagram 4 Hitung/ Calculate (a) laju purata, dalam m s–1, dalam tempoh 10 s pertama. the average speed, in m s–1, in the first 10 s. [3 markah/ marks] Jumlah jarak/ Total distance: [ 1 2 (4 + 8)(8)] + [8(10 − 8)] = 48 + 16 = 64 m Laju purata/ Average speed = 64 10 = 6.4 m s–1 (b) nilai t, jika jarak yang dilalui dalam masa 8 saat pertama adalah dua kali jarak yang dilalui dengan laju seragam. the value of t, if the distance travelled by the car in the first 8 seconds is twice the distance travelled with uniform speed. [3 markah/ marks] 1 2 (4 + 8)(8) = 2(8)(t − 8) 48 = 16t − 128 16t = 176 t = 11 s (c) nilai v, jika pecutan dalam masa 2 saat terakhir ialah 2 m s–2. the value of v, if the acceleration in the last 2 seconds is 2 m s–2. [2 markah/ marks] v – 8 13 – 11 = 2 v – 8 = 4 v = 12 m s CONTOH –1

100 Praktis Intensif 8.1 Serakan Dispersion Buku Teks m/s 212 – 218 1 Tentukan nilai data tertinggi dan nilai data terendah bagi setiap set data yang berikut. SP: 8.1.1 TP1 Mudah Determine the highest data value and the lowest data value for each of the following sets of data. (a) Jisim bagi 10 biji tembikai/ Mass of 10 watermelons 1 kg 1.5 kg 1.2 kg 0.9 kg 1.3 kg 2.2 kg 2 kg 1.4 kg 1.5 kg 1.3 kg Nilai tertinggi/ Highest value: 2.2 kg Nilai terendah/ Lowest value: 0.9 kg (b) Ketinggian murid dalam kelas Height of students in class 152 154 156 158 160 162 Tinggi (cm)/ Height (cm) Nilai tertinggi/ Highest value: 162 cm Nilai terendah/ Lowest value: 152 cm (c) Jualan mi dalam dua minggu Noodle sales in two weeks 1 8 2 1 2 4 3 1 5 5 7 8 8 4 0 0 3 5 Batang/ Stem Daun/ Leaf Kekunci 1 | 8 bermaksud 18 mangkuk Key 1 | 8 means 18 bowls Nilai tertinggi/ Highest value: 45 mangkuk/ bowls Nilai terendah/ Lowest value: 18 mangkuk/ bowls Nilai tertinggi/ Highest value: 20 Nilai terendah/ Lowest value: 8 12 8 12 17 20 11 18 11 19 16 Contoh 2 Bina satu plot titik bagi cerapan yang berikut. SP: 8.1.1 TP1 Mudah Construct a dot plot for the following observations. Jadual berikut menunjukkan poin yang diperoleh sekumpulan murid dalam suatu pertandingan. The following table shows the points obtained by a group of students in a competition. 3 5 8 3 5 3 5 8 5 9 3 5 8 5 3 10 3 4 5 6 7 8 9 10 Poin yang diperoleh Points obtained Jadual berikut menunjukkan masa yang diambil, dalam jam, oleh murid dalam Kelas 2 Mawar untuk menyiapkan kerja rumah. The following table shows the time taken, in hour, for students in Class 2 Mawar to complete their homeworks. 1 2 1.5 3 1.5 2 1 2.5 1 1.5 1.5 3 1.5 3 2.5 1 1.5 2 2.5 3 Masa yang diambil (jam) Time taken (hour) Contoh Info Digital 8.1 Sukatan Serakan Data Tak Terkumpul Measures of Dispersion for Ungrouped Data 8 Bab CONTOH

Matematik Tingkatan 4 Bab 8 101 3 Bina satu plot batang-dan-daun bagi cerapan yang berikut. SP: 8.1.1 TP1 Mudah Construct a stem-and-leaf plot for the following observations. Jadual berikut menunjukkan markah yang diperoleh murid dalam ujian Matematik. The following table shows the marks obtained by students in a Mathematics test. 35 67 36 50 49 48 82 75 3 5 6 4 8 9 5 0 6 7 7 5 8 2 Batang Stem Daun Leaf Kekunci: 3 | 5 bermaksud 35 markah Key: 3 | 5 means 35 marks Jadual berikut menunjukkan mata ganjaran yang dikumpulkan oleh beberapa orang pengunjung di sebuah pusat beli-belah. The following table shows the reward points collected by a few visitors in a shopping mall. 4 10 23 34 42 46 26 48 12 15 8 36 0 4 8 1 0 2 5 2 3 6 3 4 6 4 2 6 8 Batang Stem Daun Leaf Kekunci: 0 | 4 bermaksud 4 mata ganjaran Key: 0 | 4 means 4 reward points Contoh 4 Selesaikan masalah berikut. SP: 8.1.2 TP2 TP3 Sederhana Solve the following problems. Data berikut menunjukkan jisim, dalam kg, bagi dua longgok durian. The following data shows the mass, in kg, of two piles of durians. 1.5 4 2.5 3 3.5 2.5 2.5 1 1 3.5 3 3.5 2 1.5 4 1 4 3.5 3 2 Longgok A/ Pile A 4.5 3 4 3.5 5 2.5 4 5 4.5 3.5 4.5 3.5 3 4 5 2 3 2.5 5 3.5 Longgok B/ Pile B (i) Wakilkan kedua-dua taburan data dalam plot titik. Represent both distribution of data in dot plot. (ii) Huraikan secara ringkas taburan jisim durian bagi kedua-dua longgokan. Describe briefly the mass distribution of durians in both piles. (iii) Harga bagi kedua-dua longgokan durian adalah sama. Encik Wong hendak membeli salah satu longgok durian, yang manakah lebih berbaloi? Justifikasikan. The prices of both piles of durians are the same. Mr Wong wants to buy one of the piles, which one is more worthy? Justify. (ii) Taburan jisim durian bagi kedua-dua longgokan adalah sama kerana beza cerapan adalah sama, iaitu 3 kg. The mass distribution of durians for both piles are the same because the difference in observations is the same, which is 3 kg. (iii) Longgokan B, kerana taburan jisim durian pencong ke kiri plot titik. Pile B, because the mass distribute of durians skewed to the left of the dot plot. (i) 1 1 1.5 1.5 2.5 2.5 3.5 3.5 4.5 4.5 2 2 3 3 4 4 5 5 Jisim durian longgokan A (kg) Mass of durians of pile A (kg) Jisim durian longgokan B (kg) Mass of durians of pile B (kg) Contoh CONTOH

Matematik Tingkatan 4 Bab 8 102 Data berikut menunjukkan umur pekerja dalam dua buah kilang. The following data shows the age of workers in two factories. 47 30 51 35 27 24 34 22 44 49 38 55 48 35 Kilang P Factory P 46 49 32 50 36 45 35 42 47 29 36 50 49 52 Kilang Q Factory Q (a) Wakilkan taburan data di atas dalam plot batang-dan-daun. Represent the distribution of the above data in stem-and-leaf plot. (b) Data yang manakah mempunyai serakan yang lebih besar? Justifikasikan. Which data has a bigger dispersion? Justify. (c) Sekiranya kemahiran pekerja adalah berkadaran dengan umurnya, kilang yang manakah mempunyai prestasi yang lebih tinggi? If the skill of a worker is proportional to his age, which factory has higher performance? (a) 7 4 2 2 9 8 5 5 4 0 3 2 5 6 6 9 8 7 4 4 2 5 6 7 9 9 5 1 5 0 0 2 Kilang P Factory P Kilang Q Factory Q Kekunci: 2 | 2 | 9 bermaksud 22 tahun dan 29 tahun Key: 2 | 2 | 9 means 22 years old and 29 years old (b) Kilang P/ Factory P: Nilai tertinggi/ Highest value = 55 tahun/ years old Nilai terendah/ Lowest value = 22 tahun/ years old Beza/ Difference: 55 – 22 = 33 tahun/ years old Kilang Q/ Factory Q: Nilai tertinggi/ Highest value = 52 tahun/ years old Nilai terendah/ Lowest value = 29 tahun/ years old Beza/ Difference: 52 – 29 = 23 tahun/ years old ∴Kilang P mempunyai serakan lebih besar kerana beza cerapan datanya adalah lebih tinggi. Factory P has higher dispersion because the difference in the observation data is higher. (c) Kilang Q mempunyai prestasi yang lebih tinggi kerana kilang tersebut mempunyai pekerja dengan umur yang lebih tinggi dan lebih berkemahiran. Factory Q has higher performance because the factory has workers with older age and more skilled. CONTOH

Matematik Tingkatan 4 Bab 8 103 8.2 Sukatan Serakan Measures of Dispersion Buku Teks m/s 219 – 237 1 Tentukan julat bagi setiap set data yang berikut. SP: 8.2.1 TP2 Mudah Determine the range for each of the following sets of data. (a) 5, 9, 20, 17, 21, 13, 6 Nilai tertinggi/ Highest value = 21 Nilai terendah/ Lowest value = 5 Julat/ Range: 21 – 5 = 16 (b) Skor permainan bagi sekumpulan pemain The game score of a group of players Skor Score 2 4 6 8 10 Bilangan pemain Number of players 2 5 3 7 1 Nilai tertinggi/ Highest value = 10 Nilai terendah/ Lowest value = 2 Julat/ Range: 10 – 2 = 8 (c) Markah peserta dalam suatu pertandingan Marks of participants in a competition Markah Marks 1 4 5 8 9 12 Bilangan peserta Number of participants 1 2 3 4 3 1 Nilai tertinggi/ Highest value = 12 Nilai terendah/ Lowest value = 1 Julat/ Range: 12 – 1 = 11 10, 14, 23, 19, 33, 20 Nilai tertinggi/ Highest value = 33 Nilai terendah/ Lowest value = 10 Julat/ Range: 33 – 10 = 23 Contoh 2 Cari julat antara kuartil bagi setiap set data yang berikut. SP: 8.2.1 TP2 Mudah Find the interquartile range for each of the following sets of data. (a) 24, 17, 25, 30, 7, 42, 50, 37 Susun semula/ Rearrange: Median Q1 Q3 7, 17, 24, 25, 30, 37, 42, 50 Julat antara kuartil/ Interquartile range: 37 + 42 2 – 17 + 24 2 = 39.5 – 20.5 = 19 (b) 17, 17, 18, 18, 18, 18, 19, 19, 20, 20, 20, 21, 22 Median Q1 Q3 17, 17, 18, 18, 18, 18, 19, 19, 20, 20, 20, 21, 22 Julat antara kuartil/ Interquartile range: 20 + 20 2 – 18 + 18 2 = 20 – 18 = 2 (c) 34, 44, 51, 21, 57, 86, 23, 78, 52, 51, 69 Susun semula/ Rearrange: Median Q1 Q3 21, 23, 34, 44, 51, 51, 52, 57, 69, 78, 86 Julat antara kuartil/ Interquartile range: 69 – 34 = 35 14, 10, 6, 13, 20, 17 Susun semula/ Rearrange: 6, 10, 13, 14, 17, 20 Median Q1 Q3 Julat antara kuartil/ Interquartile range: 17 – 10 = 7 Contoh Penggunaan Kalkulator Info Digital 8.2 CONTOH

Matematik Tingkatan 4 Bab 8 104 3 Cari julat antara kuartil bagi setiap set data yang berikut. SP: 8.2.1 TP2 TP3 Sederhana Find the interquartile range for each of the following sets of data. (a) Bilangan buku yang dibaca oleh sekumpulan murid The number of books read by a group of students Bilangan buku Number of books 5 6 7 8 9 Bilangan murid Number of students 4 1 2 2 1 Kekerapan longgokan Cumulative frequency 4 5 7 9 10 Q1 = cerapan ke-(10 4 )/ ( 10 4 )th observation = cerapan ke-2.5/ 2.5th observation = 5 Q3 = cerapan ke-[ 3(10) 4 ]/ [ 3(10) 4 ]th observation = cerapan ke-7.5/ 7.5th observation = 7 + 8 2 = 7.5 Julat antara kuartil/ Interquartile range: 7.5 – 5 = 2.5 (b) Bilangan guli dalam 28 buah beg Number of marbles in 28 bags Bilangan guli Number of marbles 3 4 5 6 7 Bilangan beg Number of bags 5 4 6 8 5 Kekerapan longgokan Cumulative frequency 5 9 15 23 28 Q1 = cerapan ke-(28 4 )/ ( 28 4 )th observation = cerapan ke-7/ 7th observation = 4 Q3 = cerapan ke-[ 3(28) 4 ]/ [ 3(28) 4 ]th observation = cerapan ke-21/ 21th observation = 6 Julat antara kuartil/ Interquartile range: 6 – 4 = 2 (c) Umur ahli di dalam sebuah persatuan The age of members in a society Umur Age 14 15 17 21 23 25 Bilangan ahli Number of members 1 4 2 1 3 1 Kekerapan longgokan Cumulative frequency 1 5 7 8 11 12 Q1 = cerapan ke-(12 4 )/ ( 12 4 )th observation = cerapan ke-3/ 3th observation = 15 Q3 = cerapan ke-[ 3(12) 4 ]/ [ 3(12) 4 ]th observation = cerapan ke-9/ 9th observation = 23 Julat antara kuartil/ Interquartile range: 23 – 15 = 8 Bilangan setem yang dikumpul oleh sekumpulan murid The number of stamps collected by a group of students Bilangan setem Number of stamps 40 50 60 70 80 Bilangan murid Number of students 5 7 8 7 5 Kekerapan longgokan Cumulative frequency 5 12 20 27 32 Q1 = cerapan ke-( 32 4 )/ ( 32 4 )th observation = cerapan ke-8/ 8th observation = 50 Q3 = cerapan ke-[ 3(32) 4 ]/ [ 3(32) 4 ]th observation = cerapan ke-24/ 24th observation = 70 Julat antara kuartil/ Interquartile range: 70 – 50 = 20 Contoh Penggunaan Kalkulator Tip Bestari Formula bagi kuartil: Formula for quartiles: Q1 = N 4 Q3 = 3N 4 CONTOH

Matematik Tingkatan 4 Bab 8 105 4 Tentukan varians dan sisihan piawai bagi setiap set data yang berikut. SP: 8.2.1 TP2 TP3 Sederhana Determine the variance and standard deviation for each of the following sets of data. (a) 110, 105, 98, 63, 87, 100, 57, 48, 52 Min/ Mean, x : 110 + 105 + 98 + 63 + 87 + 100 + 57 + 48 + 52 9 = 720 9 = 80 Varians/ Variance, σ2 : (110 – 80)2 + (105 – 80)2 + (98 – 80)2 + (63 – 80)2 + (87 – 80)2 + (100 – 80)2 + (57 – 80)2 + (48 – 80)2 + (52 – 80 )2 9 = 900 + 625 + 324 + 289 + 49 + 400 + 529 + 1 024 + 784 9 = 547.111 Sisihan piawai/ Standard deviation, σ: √547.111 = 23.39 21, 34, 44, 17, 25, 9 Min/ Mean, x: 21 + 34 + 44 + 17 + 25 + 9 6 = 150 6 = 25 Varians/ Variance, σ2 : (21 – 25)2 + (34 – 25)2 + (44 – 25)2 + (17 – 25)2 + (25 – 25)2 + (9 – 25)2 6 = 16 + 81 + 361 + 64 + 0 + 256 6 = 129.667 Sisihan piawai/ Standard deviation, σ: √129.667 = 11.387 Contoh (b) Jadual berikut menunjukkan bilangan buku yang dibaca oleh sekumpulan murid dalam satu hari. The table below shows the number of books read by a group of students in one day. Bilangan buku/ Number of books 1 2 3 4 5 6 Bilangan murid/Number of students 1 6 10 3 7 3 Min/ Mean, x: 1(1) + 6(2) + 10(3) + 3(4) + 7(5) + 3(6) 1 + 6 + 10 + 3 + 7 + 3 = 108 30 = 3.6 Varians/ Variance, σ2 : 1(1 – 3.6)2 + 6(2 – 3.6)2 + 10(3 – 3.6)2 + 3(4 – 3.6)2 + 7(5 – 3.6)2 + 3(6 – 3.6)2 30 = 6.76 + 15.36 + 3.6 + 0.48 + 13.72 + 17.28 30 = 1.907 Sisihan piawai/ Standard deviation, σ: √1.907 = 1.381 Jadual berikut menunjukkan markah murid dalam suatu kuiz Matematik. The following table shows the marks of students in a Mathematical quiz. Markah/ Marks 5 10 15 20 25 Bilangan murid/Number of students 10 4 5 4 2 Min/ Mean, x: 10(5) + 4(10) + 5(15) + 4(20) + 2(25) 10 + 4 + 5 + 4 + 2 = 295 25 = 11.8 Varians/ Variance, σ2 : 10(5 – 11.8)2 + 4(10 – 11.8)2 + 5(15 – 11.8)2 + 4(20 – 11.8)2 + 2(25 – 11.8)2 25 = 462.4 + 12.96 + 51.2 + 268.96 + 348.48 25 = 45.76 Sisihan piawai/ Standard deviation, σ: √45.76 = 6.765 Contoh Penggunaan Kalkulator CONTOH

Matematik Tingkatan 4 Bab 8 106 5 Tentukan varians dan sisihan piawai bagi setiap set data yang berikut. SP: 8.2.1 TP2 TP3 Sederhana Determine the variance and standard deviation for each of the following sets of data. (a) 23, 5, 14, 20, 18, 25 x 23 5 14 20 18 25 Σx = 105 x2 529 25 196 400 324 625 Σx2 = 2 099 Min/ Mean, x: 105 6 = 17.5 Varians/ Variance, σ2 : 2 099 6 – 17.52 = 43.583 Sisihan piawai/ Standard deviation, σ: √43.583 = 6.602 (b) Rajah berikut menunjukkan panjang, dalam cm, batang kayu. The following diagram shows the length, in cm, of wooden sticks. 10 22 30 17 18 25 32 8 16 22 x x2 10 100 22 484 30 900 17 289 18 324 25 625 32 1 024 8 64 16 256 22 484 Σx = 200 Σx2 = 4 550 Min/ Mean, x: 200 10 = 20 Varians/ Variance, σ2 : 4 550 10 – 202 = 455 – 400 = 55 Sisihan piawai, σ: Standard deviation √55 = 7.416 (c) Rajah berikut menunjukkan jisim, dalam kg, bagi sekumpulan murid. The following diagram shows the mass, in kg of a group of students. 60 41 53 44 54 45 39 42 x x2 60 3 600 41 1 681 53 2 809 44 1 936 54 2 916 45 2 025 39 1 521 42 1 764 Σx = 378 Σx2 = 18 252 Min/ Mean, x: 378 8 = 47.25 Varians/ Variance, σ2 : 18 252 8 – 47.252 = 2 281.5 – 2 232.5625 = 48.9375 Sisihan piawai, σ: Standard deviation √48.9375 = 6.996 (d) Bilangan kertas dalam beberapa kotak Number of papers in several boxes Bilangan kertas/ Number of papers 50 100 150 200 Bilangan kotak/Number of boxes 4 5 3 2 x f x2 fx fx2 50 4 2 500 200 10 000 100 5 10 000 500 50 000 150 3 22 500 450 67 500 200 2 40 000 400 80 000 Σf = 14 Σfx = 1 550 Σfx2 = 207 500 Min/ Mean, x: 1 550 14 = 110.714 Varians/ Variance, σ2 : 207 500 14 – 110.7142 Sisihan piawai/ Standard deviation, σ: = 14 821.429 – 12 257.590 √2 563.839 = 50.634 = 2 563.839 10, 14, 16, 18, 12 x 10 12 14 16 18 Σx = 70 x2 100 144 196 256 324 Σx2 = 1 020 Min/ Mean, x: 70 5 = 14 Varians/ Variance, σ2 : Σx2 N – x = 1 020 5 – 142 = 204 – 196 = 8 Sisihan piawai/ Standard deviation, σ: √8 = 2.828 Contoh CONTOH

Matematik Tingkatan 4 Bab 8 107 6 Selesaikan setiap soalan yang berikut. SP: 8.2.2 TP4 TP5 Sukar Solve each of the following questions. (a) Jadual berikut menunjukkan masa yang diambil untuk mengulang kaji oleh sekumpulan murid. The following table shows the time taken to do revision by a group of students. Masa yang diambil (jam) Time taken (hour) 2 4 6 8 10 12 Bilangan murid Number of students 4 7 12 15 7 3 (i) Cari julat dan julat antara kuartil. Find the range and the interquartile range. (ii) Antara julat dan julat antara kuartil, yang manakah sesuai untuk menghuraikan serakan data di atas? Justifikasikan. Between range and interquartile range, which one is suitable to describe the dispersion of the above data? Justify. (i) Julat/ Range: 12 – 2 = 10 jam/ hours Q1 : Cerapan ke-(48 4 )/ ( 48 4 )th observation Q3 : Cerapan ke-[ 3(48) 4 ]/ [ 3(48) 4 ]th observation = Cerapan ke-12/ 12th observation = Cerapan ke-36/ 36th observation = 6 jam/ hours = 8 jam/ hours Julat antara kuartil/ Interquartile range: 8 – 6 = 2 jam/ hours (ii) Julat kerana set data tiada pencilan, oleh itu julat dapat memberi gambaran serakan yang lebih jelas. Range because the set of data does not have outlier, thus the range can give a clearer image of dispersion. (b) Jadual berikut menunjukkan masa yang diambil, dalam saat, untuk berlari 200 m oleh dua atlet. The following table shows the time taken, in seconds, to run 200 m by two athletes. 1 2 3 4 Sarah 21.22 21.34 22.04 20.55 Sylvia 20.33 21.45 21.04 20.58 (i) Hitungkan sisihan piawai bagi masa yang diambil oleh kedua-dua atlet. Calculate the standard deviation for the time taken by the two athletes. (ii) Tentukan siapa pelari dengan prestasi yang lebih konsisten. Terangkan. Determine who is the runner with more consistent performance. Explain. (iii) Jurulatih Ong telah memilih Sylvia untuk menyertai kejohanan. Terangkan pemilihan Jurulatih Ong. Coach Ong has selected Sylvia to join a competition. Explain Coach Ong’s selection. (i) Sarah: Sylvia: x: 21.22 + 21.34 + 22.04 + 20.55 4 x: 20.33 + 21.45 + 21.04 + 20.58 4 = 21.29 = 20.85 σ: 21.222 + 21.342 + 22.042 + 20.552 4 – 21.292 σ: 20.332 + 21.452 + 21.042 + 20.582 4 – 20.852 = 0.416 = 0.43 (ii) Sarah ialah pelari dengan prestasi yang lebih konsisten kerana sisihan piawai masa lariannya adalah lebih rendah daripada Sylvia. Sarah is the runner with a more consistent performance because the standard deviation of her running time is lower than Sylvia. (iii) Kerana masa larian Sylvia adalah lebih pendek dan prestasinya adalah lebih kurang sama dengan Sarah. Because Sylvia’s running time is shorter and her performance is about the same as Sarah. CONTOH

Matematik Tingkatan 4 Bab 8 108 7 Bina satu plot kotak bagi setiap set data yang berikut. SP: 8.2.3 TP4 Sederhana Construct a box plot for each of the following sets of data. Contoh (a) 23, 40, 32, 29, 36, 47, 50 Susun semula/ Rearrange: 23, 29, 32, 36, 40, 47, 50 Q1 Q3 Median 25 30 35 40 45 50 (b) Jisim, dalam kg, murid di sebuah kelas The mass, in kg, of students in a class 47 39 55 50 62 43 49 43 44 48 51 66 48 42 53 44 52 59 39, 42, 43, 43, 44, 44, 47, 48, 48, 49, 50, 51, 52, 53, 55, 59, 62, 66 Bilangan cerapan genap Even number of observations 9 cerapan 9 observations 9 cerapan 9 observations Median: 48 + 49 2 = 48.5 Q1 Q3 Q1 : Cerapan ke-5/ 5th observation = 44 Q3 : Cerapan ke-14/ 14th observation = 53 40 45 50 55 60 65 (c) Jualan surat khabar dalam Februari The sales of newspapers in February 8 7 7 7 8 8 9 0 0 1 3 5 5 9 10 0 3 4 4 4 6 6 7 8 9 11 1 3 3 5 6 7 Batang Stem Daun Leaf Kekunci: 8 | 7 bermaksud 87 naskhah Key: 8 | 7 means 87 newspapers Median = Purata cerapan ke-14 dan ke-15 Median = Average of 14th and 15th observations = 103 + 104 2 = 103.5 28 cerapan (bilangan genap) 28 observations (even number) 14 cerapan 14 observations 14 cerapan 14 observations Q1 : Purata cerapan ke-7 dan ke-8 = Average of 7th and 8th observations = 90 + 91 2 = 90.5 Q3 : Purata cerapan ke-21 dan ke-22 = Average of 21st and 22nd observations = 108 + 109 2 = 108.5 90 95 100 105 110 115 10, 12, 8, 6, 23, 19 Susun semula/ Rearrange: 6, 8, 10, 12, 19, 23 Min Q Max 1 Q3 Median = 10 +12 2 = 11 5 10 15 20 CONTOH

Matematik Tingkatan 4 Bab 8 109 8 Bagi setiap plot kotak yang berikut, nyatakan SP: 8.2.3 TP3 Mudah For each of the following box plot, state (i) nilai minimum dan nilai maksimum, (ii) julat, minimum value and maximum value, range, (iii) kuartil bawah, median dan kuartil atas, dan (iv) julat antara kuartil. lower quartile, median and upper quartile, and interquartile range. (a) 20 30 40 50 60 70 (i) Nilai minimum/ Minimum value = 20 Nilai maksimum/ Maximum value = 68 (ii) Julat/ Range: 68 – 20 = 48 (iii) Kuartil bawah/ Lower quartile = 34 Median/ Median = 42 Kuartil atas/ Upper quartile = 62 (iv) Julat antara kuartil/ Interquartile range: 62 – 34 = 28 (b) 6 8 10 12 14 16 (i) Nilai minimum/ Minimum value = 5.6 Nilai maksimum/ Maximum value = 15.2 (ii) Julat/ Range: 15.2 – 5.6 = 9.6 (iii) Kuartil bawah/ Lower quartile = 9.2 Median/ Median = 11.2 Kuartil atas/ Upper quartile = 12.4 (iv) Julat antara kuartil/ Interquartile range: 12.4 – 9.2 = 3.2 9 Selesaikan setiap yang berikut. SP: 8.2.4 TP4 TP5 Sukar Solve each of the following. Rajah berikut menunjukkan markah murid dalam ujian Matematik. The following diagram shows the students’ marks on a Mathematics test. 45 56 48 76 65 64 49 55 73 52 (i) Cari julat. Find the range. (ii) Cari julat baharu sekiranya semua markah murid ditambah 5. Find the new range if all the students’ marks are added 5. (iii) Nyatakan perubahan antara sukatan serakan asal dan sukatan serakan baharu. State the changes between the original measure of dispersion and the new measure of dispersion. (i) Nilai maksimum/ Maximum value = 76 Nilai minimum/ Minimum value = 45 Julat/ Range: 76 – 45 = 31 (ii) Nilai maksimum baharu/ New maximum value: 76 + 5 = 81 Nilai minimum baharu/ New minimum value: 45 + 5 = 50 Julat baharu/ New range: 81 – 50 = 31 (iii) Tiada perubahan, julat kekal sama iaitu 31. No change, the range remains the same which is 31. Contoh CONTOH

Matematik Tingkatan 4 Bab 8 110 (a) Plot batang-dan-daun berikut menunjukkan jisim, dalam kg, tin aluminium yang dikumpul. The stem-and-leaf plot shows the mass, in kg, of the aluminium cans collected. 2 6 7 7 9 3 0 1 1 4 5 6 4 2 3 4 4 5 6 6 8 5 3 5 Batang Stem Daun Leaf Kekunci: 2 | 6 bermaksud 26 kg Key: 2 | 6 means 26 kg 28 ^ 33 ^ 44 ^ 68 ^ (i) Cari julat dan julat antara kuartil. Find the range and the interquartile range. (ii) Cari julat dan julat antara kuartil baharu jika 28 kg, 33 kg, 44 kg dan 68 kg ditambah ke dalam set data. Find the new range and interquartile range if 28 kg, 33 kg, 44 kg and 68 kg are added to the set of data. (iii) Nyatakan perubahan terhadap kedua-dua serakan sukatan bagi sebelum dan selepas penambahan cerapan baharu. State the changes towards both measures of dispersion for before and after the addition of new observations. (i) Julat/ Range: 55 – 26 = 29 kg (ii) Julat/ Range: 68 – 26 = 42 kg Q1 = Purata cerapan ke-5 dan ke-6 = Average of 5th and 6th observations = 30 + 31 2 = 30.5 kg Q3 = Purata cerapan ke-15 dan ke-16 = Average of 15th and 16th observations = 45 + 46 2 = 45.5 kg Julat antara kuartil/Interquartile range: 45.5 kg – 30.5 kg = 15 kg Q1 = Purata cerapan ke-6 dan ke-7 = Average of 6th and 7th observations = 31 + 31 2 = 31 kg Q3 = Purata cerapan ke-18 dan ke-19 = Average of 18th and 19th observations = 45 + 46 2 = 45.5 kg Julat antara kuartil/Interquartile range: 45.5 kg – 31 kg = 14.5 kg (iii) Julat meningkat dengan banyak kerana terdapat satu nilai ekstrem. The range increased greatly because there is an extreme value. Julat antara kuartil masih kekal lebih kurang sama, iaitu 15 kg. The interquartile range still remains almost the same, which is 15 kg. (b) Jadual berikut menunjukkan jisim, dalam kg, surat khabar yang dikumpulkan oleh penduduk. The following table shows the mass, in kg, of newspapers collected by the residents. Jisim (kg)/ Mass (kg) 4 8 12 16 20 Bilangan penduduk/ Number of residents 3 7 6 4 2 (i) Cari sisihan piawai bagi set data di atas. Find standard deviation for the above set of data. (ii) Sekiranya jisim 20 kg dikeluarkan daripada set data tersebut, cari sisihan piawai baharu. If the mass of 20 kg is removed from the set of data, find the new standard deviation. (iii) Apakah perubahan terhadap serakan data berdasarkan sisihan piawai asal dan baharu? What is the change towards the dispersion of data based on the original and new standard deviation? (i) x 4 8 12 16 20 f 3 7 6 4 2 Σf = 22 fx 12 56 72 64 40 Σfx = 244 x2 16 64 144 256 400 fx2 48 448 864 1 024 800 Σfx2 = 3 184 (ii) x 4 8 12 16 f 3 7 6 4 Σf = 20 fx 12 56 72 64 Σfx = 204 x2 16 64 144 256 fx2 48 448 864 1 024 Σfx2 = 2 384 x= 244 22 = 11.091 x= 204 20 = 10.2 Sisihan piawai/ Standard deviation: Sisihan piawai/ Standard deviation: 3 184 22 – 11.0912 = 4.66 2 384 20 – 10.22 = 3.89 (iii) Serakan data menjadi kecil selepas jisim 20 kg dikeluarkan kerana sisihan piawai baharu menjadi kecil. The dispersion of data becomes smaller after the mass of 20 kg is removed because the new standard deviation has become smaller. CONTOH

Matematik Tingkatan 4 Bab 8 111 10 Berdasarkan perwakilan grafik atau jadual berikut, tentukan perubahan yang dikenakan ke atas set data. SP: 8.2.4 Based on the following graphical representation or table, determine the changes done to the sets of data. TP4 Sederhana (a) Set data asal/ Original set of data: 2 1 3 9 9 4 3 3 5 6 8 5 0 2 Batang Stem Daun Leaf Kekunci: 2 | 1 bermaksud 21 kg Key: 2 | 1 means 21 kg Set data baharu/ New set of data: 3 9 9 4 3 3 5 6 8 5 0 2 Batang Stem Daun Leaf Kekunci: 3 | 9 bermaksud 39 kg Key: 3 | 9 means 39 kg Pencilan 21 kg dikeluarkan. The outlier 21 kg is removed. (b) Jadual berikut menunjukkan min dan sisihan piawai yang diperoleh daripada keputusan suatu eksperimen. The following table shows the mean and standard deviation obtained from the results of an experiment. Bilangan cerapan Number of observations Min Mean Sisihan piawai Standard deviation Set data asal/ Original set of data: 20 55.5 0.78 Set data baharu/ New set of data: 21 57.8 1.23 Satu nilai yang jauh daripada min ditambah ke dalam set data tersebut. A value which is far from the mean was added to the set of data. 9 9 10 10 9.5 9.5 10.5 10.5 11 11 Jisim tembikai (kg) Mass of watermelons (kg) Jisim tembikai (kg) Mass of watermelons (kg) Set data asal: Original set of data: Set data baharu: New set of data: Semua cerapan ditambah 0.5 kg. All observations are added 0.5 kg. Jadual berikut menunjukkan min dan sisihan piawai daripada keputusan bagi satu eksperimen. The following table shows the mean and standard deviation from the results of an experiment. Bilangan cerapan Number of observations Min Mean Sisihan piawai Standard deviation Set data asal/ Original set of data: 10 4.5 1.5 Set data baharu/ New set of data: 8 4.2 0.4 Dua nilai yang jauh daripada min dikeluarkan daripada set data tersebut. Two values which are far from the mean were removed from the set of data. Contoh CONTOH

Matematik Tingkatan 4 Bab 8 112 11 Selesaikan setiap masalah yang berikut. SP: 8.2.5 TP5 Sukar Solve each of the following problems. (a) Plot batang-dan-daun di bawah menunjukkan markah ujian Matematik yang diperoleh murid-murid cemerlang di dalam dua buah kelas. The stem-and-leaf plot below shows the marks of Mathematics test obtained by outstanding students in two classes. 8 7 6 6 9 8 7 7 4 0 7 1 5 6 6 9 8 7 3 8 4 5 6 7 9 9 7 2 9 5 6 8 Kelas 2 Melati Class 2 Melati Kelas 2 Ros Class 2 Ros Kekunci: 6 | 6 | 9 bermaksud 66 markah dan 69 markah Key: 6 | 6 | 9 means 66 marks and 69 marks (i) Hitungkan min dan sisihan piawai markah bagi kedua-dua kelas. Calculate the mean and standard deviation of the marks for both classes. (ii) Huraikan secara ringkas markah murid-murid dari kedua-dua kelas berdasarkan sisihan piawai. Briefly describe the marks of the students from both classes based on the standard deviation. (iii) Kelas yang paling cemerlang akan diberi ganjaran. Antara kedua-dua kelas tersebut, yang manakah akan mendapat ganjaran? Terangkan. The most outstanding class will be rewarded. Between the two classes, which one will be rewarded? Explain. (i) Kelas 2 Melati/ Class 2 Melati Min/ Mean: 66 + 67 + 68 + 70 + 74 + 77 + 77 + 78 + 83 + 87 + 88 + 89 + 92 + 97 14 = 79.5 Sisihan piawai/ Standard deviation: 662 + 672 + 682 + 702 + 742 + 772 + 772 + 782 + 832 + 872 + 882 + 892 + 922 + 972 14 – 79.52 = 89 783 14 – 6 320.25 = √92.821 = 9.63 Kelas 2 Ros/ Class 2 Ros Min/ Mean: 69 + 71 + 75 + 76 + 76 + 84 + 85 + 86 + 87 + 89 + 89 + 95 + 96 + 98 14 = 84 Sisihan piawai/ Standard deviation: 692 + 712 + 752 + 762 + 762 + 842 + 852 + 862 + 872 + 892 + 892 + 952 + 962 + 98 2 14 – 842 = 99 912 14 – 7 056 = √80.571 = 8.98 (ii) Markah murid di Kelas 2 Melati adalah lebih berserak manakala markah murid di Kelas 2 Ros adalah lebih bertumpu kepada minnya. The marks of students in Class 2 Melati is more dispersed whereas the marks of students in Class 2 Ros is more converge towards its mean. (iii) Kelas 2 Ros kerana kebanyakan muridnya memperoleh markah yang lebih tinggi daripada Kelas 2 Melati. Class 2 Ros because most of its students obtained higher marks than Class 2 Melati. CONTOH

Matematik Tingkatan 4 Bab 8 113 (b) Plot kotak berikut menunjukkan jangka hayat, dalam hari, bagi tiga jenis mentol. The following box plot shows the life spans, in days, of three types of bulbs. 50 55 60 65 70 75 Jenis A/ Type A Jenis B/ Type B Jenis C/ Type C Jangka hayat (hari) Life span (days) (i) Nyatakan julat dan julat antara kuartil jangka hayat bagi ketiga-tiga jenis mentol. State the range and interquartile range of the life spans for the three types of bulbs. (ii) Encik Rosdi hendak membeli mentol jenis A tetapi penjual mencadangkan mentol jenis C. Wajarkan cadangan penjual. Mr Rosdi wants to buy type A bulb but the seller suggested type C bulb. Justify the seller’s suggestion. (iii) Sekiranya harga ketiga-tiga jenis mentol adalah sama, pembelian jenis mentol yang manakah adalah lebih berbaloi? Terangkan dengan sukatan kecenderungan memusat dan sukatan serakan yang sesuai. If the prices for the three types of bulbs are the same, the purchase of which type of bulb is more worthy? Explain using a suitable measure of central tendency and measure of dispersion. (i) Jenis mentol Type of bulb Julat Range Julat antara kuartil Interquartile range A 75 – 50 = 25 hari/ days 71 – 57 = 14 hari/ days B 77 – 62 = 15 hari/ days 74 – 66 = 8 hari/ days C 75 – 57 = 18 hari/ days 72 – 60 = 12 hari/ days (ii) Taburan jangka hayat bagi mentol jenis C adalah lebih bertumpu ke kanan berbanding dengan mentol jenis A, maka penggunaan mentol jenis C adalah lebih tahan lama. The distribution of life span for type C bulb converged more to the right compared to type A bulb, thus the usage of type C bulb lasts longer. (iii) Mentol jenis B kerana antara tiga jenis mentol itu, median jangka hayat bagi mentol jenis B adalah paling tinggi dan taburan jangka hayatnya adalah agak konsisten dengan julat antara kuartil yang rendah. Type B bulb because among the three types of bulbs, the median of life span of type B bulb is the highest and the distribution of its life span is very consistent with short interquartile range. CONTOH

Matematik Tingkatan 4 Bab 8 114 (c) Jadual dan plot kotak berikut menunjukkan taburan ketinggian bagi 36 batang pokok durian. The following table and box plot shows the height distribution of 36 durian trees. Tinggi (m) Height (m) 35 40 45 50 Bilangan pokok Number of trees 5 13 8 10 35 40 45 50 Tinggi (m)/ Height (m) (i) Nyatakan julat dan julat antara kuartil bagi taburan tinggi pokok durian di atas. State the range and interquartile range for the height distribution of durian trees above. (ii) Cari julat dan julat antara kuartil yang baharu sekiranya cerapan 30 m, 30 m, 50 m dan 65 m dimasukkan ke dalam set data di atas. Find the new range and interquartile range if observations of 30 m, 30 m, 50 m and 65 m are added to the above set of data. (iii) Tentukan sama ada julat baharu boleh digunakan untuk mewakili taburan data tersebut. Wajarkan. Determine whether the new range can be used to represent the distribution of data. Justify. (i) Julat/ Range: 50 – 35 = 15 m Julat antara kuartil/ Interquartile range: 50 – 40 = 10 m (ii) Tinggi (m) Height (m) 30 35 40 45 50 65 Bilangan pokok Number of trees 2 5 13 8 11 1 Julat baharu/ New range: 65 – 30 = 35 m Q1 : Cerapan ke-(40 4 )/ ( 40 4 )th observation = Cerapan ke-10/ 10th observation = 40 m Q3 : Cerapan ke-[ 3(40) 4 ]/ [ 3(40) 4 ]th observation = Cerapan ke-30/ 30th observation = 50 m Julat antara kuartil baharu/ New interquartile range: 50 – 40 = 10 m (iii) Tidak, kerana wujud pencilan. No, because there is an outlier. CONTOH

Matematik Tingkatan 4 Bab 8 115 12 Selesaikan setiap soalan yang berikut. SP: 8.2.6 TP6 KBAT Sukar Solve each of the following questions. (a) Plot titik yang tidak lengkap berikut menunjukkan masa yang diambil untuk melatih tarian bagi beberapa orang penari dalam seminggu. The following incomplete dot plot shows the time taken to practice dancing for a number of dancers in one week. 15 20 25 30 35 40 Masa (jam)/ Time (hours) (i) Diberi min ialah 27.5 jam dan sisihan piawai ialah 7.331 jam. Jika Σfx2 = 16 200 jam2 , cari bilangan penari terlibat dalam set data ini. Given that the mean is 27.5 hours and the standard deviation is 7.331 hours. If Σfx2 = 16 200 hour2 , find the number of dancers involved in this set of data. (ii) Beza masa latihan antara penari yang tertinggal daripada set data tersebut ialah 5 jam. Cari masa latihan yang mungkin bagi semua penari yang tertinggal. The difference in practice time between the dancers left out from the set of data is 5 hours. Find the possible practice time for all the left-out dancers. (i) σ = Σfx2 Σf – x2 7.331 = 16 200 Σf – 27.52 53.744 = 16 200 Σf – 756.25 809.994 = 16 200 Σf Σf = 20 orang penari/ dancers (ii) Bilangan penari yang tertinggal: 20 – 18 = 2 Number of dancers left out Jumlah masa latihan bagi penari yang tertinggal: Total practice time of the left-out dancers: 2(15) + 3(20) + 6(25) + 4(30) + 35 + 2(40) + x= 20(27.5) 30 + 60 + 150 + 120 + 35 + 80 + x = 550 x = 75 75 – 5 2 = 35 jam/ hours ∴ Masa latihan bagi dua orang penari yang tertinggal ialah 35 jam dan 40 jam masingmasing. Practice times for the two left-out dancers are 35 hours and 40 hours respectively. (b) Jadual berikut menunjukkan dua set data bagi satu eksperimen yang sama. The following table shows two sets of data of the same experiment. Set Set Bilangan ulangan Number of repetition Min Mean Varians Variance I 10 40.5 1.2 II 15 44.3 1.6 Tentukan sisihan piawai bagi keputusan gabungan eksperimen tersebut. Determine the standard deviation for the combined results of the experiment. Set I: Σx N = x Σx 10 = 40.5 Σx = 405 Σx2 N – x2 = σ2 Σx2 10 – 40.52 = 1.2 Σx2 = 16 414.5 Set II: Σx N = x Σx 15 = 44.3 Σx = 664.5 Σx2 N – x 2 = σ2 Σx2 15 – 44.32 = 1.6 Σx2 = 29 461.35 NI + NII = 10 + 15 = 25 ΣxI + ΣxII = 405 + 664.5 = 1 069.5 Σx2 I + Σx2 II = 16 414.5 + 29 461.35 = 45 875.85 Min baharu/ New mean: 1 069.5 25 = 42.78 Sisihan piawai/ Standard deviation: 45 875.85 25 – 42.782 = √4.9056 = 2.215 CONTOH

Matematik Tingkatan 4 Bab 8 116 (c) Jadual berikut menunjukkan masa yang diambil, dalam minit, untuk menjawab satu soalan Matematik bagi sekumpulan murid. The following table shows the time taken, in minutes, to answer one Mathematical question by a group of students. Masa yang diambil (min) Time taken (min) 2 3 4 5 6 7 Bilangan murid Number of students 3 p 5 q 2 3 Diberi min = 4.44 dan sisihan piawai = 1.47. Cari nilai p dan nilai q. Given the mean = 4.44 and the standard deviation = 1.47. Find the values of p and q. Min/ Mean: 59 + 3p + 5q 13 + p + q = 4.44 59 + 3p + 5q = 57.72 + 4.44p + 4.44q 1.44p = 1.28 + 0.56q p = 1.28 + 0.56q 1.44 ………① Sisihan piawai/ Standard deviation: 311 + 9p + 25q 13 + p + q – 4.442 = 1.47 311 + 9p + 25q 13 + p + q = 2.16 + 19.71 311 + 9p + 25q = 21.87(13 + p + q) 311 + 9p + 25q = 284.31 + 21.87p + 21.87q 12.87p – 3.13q = 26.69 ………② x f fx x2 fx2 2 3 6 4 12 3 p 3p 9 9p 4 5 20 16 80 5 q 5q 25 25q 6 2 12 36 72 7 3 21 49 147 Σf = 13 + p + q Σfx = 59 + 3p + 5q Σfx2 = 311 + 9p + 25q ∴ p = 4, q = 8. Gantikan ① ke dalam ②: Replace ① into ② 12.87(1.28 + 0.56q 1.44 ) – 3.13q = 26.69 11.44 + 5q – 3.13q = 26.69 1.87q = 15.25 q = 8.16 ≈ 8 Apabila q = 8/ When q = 8, p = 1.28 + 0.56(8) 1.44 = 4 CONTOH

Matematik Tingkatan 4 Bab 8 117 Kertas 1 Jawab semua soalan./ Answer all questions. 1 Rajah 1 menunjukkan bilangan ahli keluarga dalam beberapa buah keluarga. Diagram 1 shows the number of family members in a few families. 5, 8, p, q, 4, 5 Rajah 1/ Diagram 1 Diberi min bagi set data tersebut ialah 35 6 . Hitung nilai p + q. Given the mean for the set of data is 35 6 . Calculate the value of p + q. A 10 B 11 C 12 D 13 2 Rajah 2 menunjukkan satu set data. Diagram 2 shows a set of data. 100, 96, 42, 87, 50, 60 Rajah 2/ Diagram 2 Hitung varians bagi set data dalam Rajah 2. Calculate the variance for the set of data in Diagram 2. A 518.58 B 435 C 22.73 D 16.25 3 Diberi hasil tambah bagi enam nombor ialah 90 dan variansnya ialah 15. Hitung hasil tambah kuasa dua bagi enam nombor tersebut. Given the sum of six numbers is 90 and its variance is 15. Calculate the sum of square of the six numbers. A 160 B 240 C 1 440 D 2 400 4 Varians bagi satu set data yang mengandungi lima nombor ialah 30. Cari varians baharu sekiranya setiap nombor dalam set data itu ditambah 2. The variance of a set of data which consists of five numbers is 30. Find the new variance if each number in the set of data is added 2. A 28 B 30 C 32 D 34 5 Jadual 1 menunjukkan bilangan anak bagi beberapa buah keluarga. Table 1 shows the number of children of a few families. Bilangan anak Number of children 0 1 2 3 4 5 6 Bilangan keluarga Number of families 1 5 4 2 3 1 1 Jadual 1/ Table 1 Hitung sisihan piawai bagi set data dalam Jadual 1. Calculate the standard deviation for the set of data in Table 1. A 1.17 C 2.47 B 1.61 D 2.60 6 Rajah 3 menunjukkan satu set data. Diagram 3 shows a set of data. 15, 18, 8, 20, 22, 28 Rajah 3/ Diagram 3 Jika setiap nombor dalam set data tersebut didarab dengan 2, cari sisihan piawai baharu. If each number in the set of data is multiplied by 2, find the new standard deviation. A 6.16 C 18.5 B 12.32 D 37.92 7 Jadual 2 menunjukkan saiz kasut murid di dalam sebuah kelas. Table 2 shows the shoe size of the students in a class. Saiz kasut Shoe size 34 35 36 37 38 Bilangan murid Number of students 2 3 5 4 1 Jadual 2/ Table 2 Jika seorang murid yang memakai saiz kasut 42 dimasukkan ke dalam set data tersebut, tentukan sukatan serakan yang sesuai untuk mewakili taburan saiz kasut bagi kelas tersebut. If a student who wears shoe size 42 is added to the set of data, determine the suitable measure of dispersion to represent the distribution of shoe size of the class. A Julat Range B Varians Variance C Sisihan piawai Standard deviation D Julat antara kuartil Interquartile range K B A T Praktis Kendiri CONTOH

Matematik Tingkatan 4 Bab 8 118 Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 (a) Cari julat antara kuartil bagi setiap set data yang berikut. Find the interquartile range for each of the following sets of data sets. (i) 59, 32, 43, 56, 78, 66, 88, 35, 44 (ii) 0.4, 1.2, 2.2, 0.2, 1.35, 2.4, 3.1, 1.5, 0.8, 2.1, 2.4, 1.6 [2 markah/ marks] (i) Susun semula/ Rearrange: Q1 Q Median 3 32, 35, 43, 44, 56, 59, 66, 78, 88 Julat antara kuartil/ Interquartile range: 66 + 78 2 – 35 + 43 2 = 72 – 39 = 33 (ii) Susun semula/ Rearrange: Q1 Q Median 3 0.2, 0.4, 0.8, 1.2, 1.35, 1.5, 1.6, 2.1, 2.2, 2.4, 2.4, 3.1 Julat antara kuartil/ Interquartile range: 2.2 + 2.4 2 – 0.8 + 1.2 2 = 2.3 – 1.0 = 1.3 (b) Cari sisihan piawai bagi set data 5, 3, 7, 8, 8, 6, 7, 5, 7, 9 dan seterusnya, komen tentang serakan data. Find the standard deviation for the set of data 5, 3, 7, 8, 8, 6, 7, 5, 7, 9 and hence, comment on the dispersion of the data. [3 markah/ marks] x : 5 + 3 + 7 + 8 + 8 + 6 + 7 + 5 + 7 + 9 10 = 65 10 = 6.5 Σx2 N : 52 + 32 + 72 + 82 + 82 + 62 + 72 + 52 + 72 + 92 10 = 451 10 = 45.1 Sisihan piawai/ Standard deviation: σ2 = 45.1 – 6.52 = 45.1 – 42.25 = 2.85 σ = 1.688 ∴Dengan nilai sisihan piawai yang besar, cerapan tidak berserak dekat min. With a big value of standard deviation, the data are not distributed close to the mean. 2 Rajah 1 menunjukkan jisim, dalam kg, bagi 10 biji tembikai. Diagram 1 shows the mass, in kg, of 10 watermelons. 10 20 11 22 14 17 16 21 20 15 Rajah 1/ Diagram 1 (a) Cari julat dan julat antara kuartil bagi data di atas. Find the range and the interquartile range for the data above. [2 markah/ marks] Julat/ Range: 22 – 10 = 12 kg Susun semula/ Rearrange: Q1 Q Median 3 10, 11, 14, 15, 16, 17, 20, 20, 21, 22 Julat antara kuartil/ Interquartile range: 20 – 14 = 6 kg (b) Lakarkan satu plot kotak untuk mewakili data di atas. Sketch a box plot to represent the data above. [2 markah/ marks] Jisim/ Mass (kg) 10 14 16.5 20 22 Bahagian B/ Section B 3 Rajah 2 menunjukkan pendapatan harian, dalam RM, bagi beberapa orang pekerja di sebuah kilang. Diagram 2 shows the daily incomes, in RM, of a few workers at a factory. 35 40 48 54 44 50 42 65 38 50 55 39 Rajah 2/ Diagram 2 (a) Wakilkan pendapatan harian pekerja tersebut dalam sebuah plot batang-dan-daun. Represent the daily incomes of the workers in a stem-and-leaf plot. [3 markah/ marks] 3 5 8 9 4 0 2 4 8 5 0 0 4 5 6 5 Batang Stem Daun Leaf Kekunci: 3 | 5 bermaksud RM35 Key: 3 | 5 means RM35 CONTOH

Matematik Tingkatan 4 Bab 8 119 (b) Hitung julat antara kuartil bagi set data tersebut. Calculate the interquartile range for the set of data. [2 markah/ marks] 35, 38, 39, 40, 42, 44, 48, 50, 50, 54, 55, 65 Q Median 1 Q3 Julat antara kuartil/ Interquartile range: 50 + 54 2 – 39 + 40 2 = 52 – 39.5 = 12.5 (c) Seterusnya, binakan satu plot kotak untuk mewakili pendapatan harian pekerja di kilang tersebut. Hence, construct a box plot to represent the daily incomes of the workers at the factory. [3 markah/ marks] 35 40 45 50 55 60 65 Pendapatan harian (RM) Daily income (RM) 4 Jadual 1 menunjukkan masa yang diambil untuk menjawab semua soalan dalam ujian Matematik oleh murid-murid di sebuah kelas. Table 1 shows the time taken to answer all the questions in a Mathematics test by the students in a class. Masa yang diambil (min) Time taken (min) Bilangan murid Number of students 40 4 45 6 50 8 55 7 60 15 Jadual 1/ Table 1 (a) Hitung varians dan sisihan piawai bagi masa yang diambil oleh murid-murid. Calculate the variance and standard deviation for the time taken by the students. [3 markah/ marks] x f fx x2 fx2 40 4 160 1 600 6 400 45 6 270 2 025 12 150 50 8 400 2 500 20 000 55 7 385 3 025 21 175 60 15 900 3 600 54 000 Σf = 40 Σfx = 2 115 Σfx2 = 113 725 x = Σfx Σf = 2 115 40 = 52.875 (b) Cari varians dan sisihan piawai baharu sekiranya masa yang diambil oleh murid-murid adalah berganda dua. Find the new variance and standard deviation if the time taken by the students is doubled. [3 markah/ marks] x f fx x2 fx2 80 4 320 6 400 25 600 90 6 540 8 100 48 600 100 8 800 10 000 80 000 110 7 770 12 100 84 700 120 15 1 800 14 400 216 000 Σf = 40 Σfx = 4 230 Σfx2 = 454 900 x = Σfx Σf = 4 230 40 = 105.75 (c) Tentukan kesan perubahan terhadap sukatan serakan asal dan baharu bagi masa yang diambil. Determine the effect of change in the original and new measures of dispersion for the time taken. [1 markah/ mark] σa : σb = 6.882 : 13.764 = 1 : 2 Apabila setiap cerapan dalam set data diganda dua, sisihan piawai baharu juga diganda dua. When each observation in the set of data is doubled, the standard deviation is also doubled. σ2 = Σfx2 Σf – x2 = 113 725 40 – 52.8752 = 2 843.125 – 2 795.766 = 47.359 σ = √47.359 = 6.882 σ2 = Σfx2 Σf – x2 = 454 900 40 – 105.752 = 11 372.5 – 11 183.063 = 189.437 σ = √189.437 = 13.764 CONTOH

Matematik Tingkatan 4 Bab 8 120 5 Jadual 2 menunjukkan prestasi dua orang atlet dalam acara larian 200 m. Table 2 shows the performance of two athletes in the 200 m sprint. 1 2 3 4 Ravi 22.30 s 23.20 s 21.56 s 22.15 s Anuar 23.50 s 21.30 s 22.43 s 22.50 s Jadual 2/ Table 2 (a) Antara Ravi dan Anuar, siapakah mempunyai masa larian yang lebih konsisten? Wajarkan. Between Ravi and Anuar, who has a more consistent running time? Justify. [5 markah/ marks] Ravi: Min/ Mean: 22.3 + 23.2 + 21.56 + 22.15 4 = 22.3025 s Sisihan piawai/ Standard deviation: 22.32 + 23.22 + 21.562 + 22.152 4 – 22.30252 = √0.345 = 0.5874 Anuar: Min/ Mean: 23.5 + 21.3 + 22.43 + 22.5 4 = 22.4325 s Sisihan piawai/ Standard deviation: 23.52 + 21.32 + 22.432 + 22.52 4 – 22.43252 = √0.6067 = 0.7789 ∴ Ravi, kerana sisihan piawai masa lariannya adalah lebih rendah. Ravi, because the standard deviation of his running time is lower. (b) Jika rekod masa yang paling panjang bagi keduadua atlet dikeluarkan daripada set data, adakah prestasi Anuar lebih baik berbanding dengan Ravi? Terangkan jawapan anda. If the longest time records for both athletes are removed from the set of data, will Anuar’s performance be better than Ravi? Explain your answer. [5 markah/ marks] Ravi: Min/ Mean: 22.3 + 21.56 + 22.15 3 = 66.01 3 = 22.003 s Sisihan piawai/ Standard deviation: 22.32 + 21.562 + 22.152 3 – (66.01 3 ) 2 = √0.102 = 0.319 Anuar: Min/ Mean: 21.3 + 22.43 + 22.5 3 = 66.23 3 = 22.077 s Sisihan piawai/ Standard deviation: 22.32 + 22.432 + 22.52 3 – (66.23 3 ) 2 = √0.3024 = 0.5499 ∴ Tidak, nilai min dan sisihan piawai Ravi masih lebih rendah daripada Anuar. No, Ravi’s mean value and standard deviation are still lower than Anuar. CONTOH

121 Praktis Intensif 9.1 Peristiwa Bergabung Combined Events Buku Teks m/s 244 – 245 1 Tulis ruang sampel bagi peristiwa bergabung berikut. SP: 9.1.1 TP1 Mudah Write down the sample space for the following combined events. (a) Melambung sekeping duit syiling adil dan sebiji dadu adil. Tossing a fair coin and a fair dice. S = {(Kepala, 1), (Kepala, 2), (Kepala, 3), (Kepala, 4), (Kepala, 5), (Kepala, 6), (Ekor, 1), (Ekor, 2), (Ekor, 3), (Ekor, 4), (Ekor, 5), (Ekor, 6)} S = {(Head, 1), (Head, 2), (Head, 3), (Head, 4), (Head, 5), (Head, 6), (Tail, 1), (Tail, 2), (Tail, 3), (Tail, 4), (Tail, 5), (Tail, 6)} (b) Empat keping kad berlabel dengan huruf "R, A, P, I" dimasukkan ke dalam sebuah kotak. Kemudian, dua keping kad dikeluarkan dari kotak tersebut secara rawak satu demi satu dengan pemulangan. Four cards labelled with letters "R, A, P, I" are put into a box. Then, two cards are drawn randomly from the box one after another with replacement. S = {(R, R), (R, A), (R, P), (R, I), (A, R), (A, A), (A, P), (A, I), (P, R), (P, A), (P, P), (P, I), (I, R), (I, A), (I, P), (I, I)} (c) Dua keping kad berlabel dengan huruf “Q, R” dan empat keping kad bernombor “5, 3, 2, 9” dimasukkan ke dalam sebuah kotak. Dua keping kad dikeluarkan dari kotak tersebut satu demi satu tanpa pemulangan. Two cards labelled with letters “Q, R” and four cards labelled with numbers “5, 3, 2, 9” are put into a box. Two cards are drawn randomly from the box one after another without replacement. S = {(Q, R), (Q, 5), (Q, 3), (Q, 2), (Q, 9), (R, Q), (R, 5), (R, 3), (R, 2), (R, 9), (5, Q), (5, R), (5, 3), (5, 2), (5, 9), (3, Q), (3, R), (3, 5), (3, 2), (3, 9), (2, Q), (2, R), (2, 5), (2, 3), (2, 9), (9, Q), (9, R), (9, 5), (9, 3), (9, 2)} (d) Lima keping kad berlabel dengan huruf "J, A, M, B, U" dimasukkan ke dalam sebuah kotak. Kemudian, dua keping kad dikeluarkan dari kotak tersebut secara rawak satu demi satu tanpa pemulangan. Five cards labelled with letters "J, A, M, B, U" are put into a box. Then, two cards are drawn randomly from the box one after another without replacement. S = {(J, A), (J, M), (J, B), (J, U), (A, J), (A, M), (A, B), (A, U), (M, J), (M, A), (M, B), (M, U), (B, J), (B, A), (B, M), (B, U), (U, J), (U, A), (U, M), (U, B)} (e) Dua helai baju dipilih secara rawak dari sebuah almari yang mengandungi sehelai baju berwarna biru, sehelai baju berwarna merah dan sehelai baju berwarna hitam. Two shirts are chosen randomly from a cupboard which contains a blue shirt, a red shirt and a black shirt. S = {(Biru, Merah), (Biru, Hitam), (Merah, Biru), (Merah, Hitam), (Hitam, Biru), (Hitam, Merah)} S = {(Blue, Red), (Blue, Black), (Red, Blue), (Red, Black), (Black, Blue), (Black, Red)} Dua keping syiling dilambung secara serentak. Two coins are tossed simultaneously. S = {(Kepala, Kepala), (Kepala, Ekor), (Ekor, Kepala), (Ekor, Ekor)} S = {(Head, Head), (Head, Tail), (Tail, Head), (Tail, Tail)} Contoh Kebarangkalian Peristiwa Bergabung Probability of Combined Events 9 Bab Info Digital 9.1 CONTOH

Matematik Tingkatan 4 Bab 9 122 2 Selesaikan masalah berikut. SP: 9.1.1 TP2 Mudah Solve the following problems. (a) Diberi bahawa A ialah set semesta, {x : 1 N x N 10, x ialah integer}, B ialah peristiwa mendapat nombor perdana dan C ialah peristiwa mendapat nombor genap. Given that A is a universal set, {x : 1 N x N 10, x is an integer}, B is an event where a prime number is chosen and C is an event where an even number is chosen. (i) Tulis ruang sampel bagi peristiwa bergabung. Write down the sample space for the combined event. (ii) Tentukan bilangan kesudahan yang mungkin bagi peristiwa bergabung tersebut. Determine the number of possible outcomes for the combined event. (i) B = {2, 3, 5, 7} C = {2, 4, 6, 8, 10} S = {(2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (3, 2), (3, 4), (3, 6), (3, 8), (3, 10), (5, 2), (5, 4), (5, 6), (5, 8), (5, 10), (7, 2), (7, 4), (7, 6), (7, 8), (7, 10)} (ii) n(S) = 20 (b) Sebiji dadu dilambung sebanyak dua kali. Diberi P ialah peristiwa mendapat nombor ganjil dan Q ialah peristiwa mendapat nombor perdana. A dice is thrown twice. Given P is an event of getting an odd number and Q is an event of getting a prime number. (i) Tulis ruang sampel bagi peristiwa bergabung. Write down the sample space for the combined event. (ii) Tentukan bilangan kesudahan yang mungkin bagi peristiwa bergabung tersebut. Determine the number of possible outcomes for the combined event. (i) P = {1, 3, 5} Q = {2, 3, 5} S = {(1, 2), (1, 3), (1, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} (ii) n(S) = 9 (c) Mandy membeli dua pemutar seperti yang ditunjukkan dalam rajah di bawah. Dia memutar kedua-dua pemutar itu pada masa yang sama. Mandy bought two spinners as shown in the diagrams below. She spins both the spinners at the same time. 4 3 1 2 1 2 Senaraikan semua kombinasi yang akan diperoleh Mandy. List out all the combinations obtained by Mandy. S = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)} Sekeping duit syiling dilambung dan satu pemutar diputar secara serentak seperti yang ditunjukkan dalam rajah di bawah. A coin is tossed and a spinner is spinned simultaneously as shown in the diagrams below. 4 3 1 2 (i) Tulis ruang sampel bagi peristiwa bergabung tersebut. Write down the sample space for the combined event. (ii) Tentukan bilangan kesudahan yang mungkin bagi peristiwa bergabung tersebut. Determine the number of possible outcomes for the combined event. (i) S = {(Kepala, 1), (Kepala, 2), (Kepala, 3), (Kepala, 4), (Ekor, 1), (Ekor, 2), (Ekor, 3), (Ekor, 4)} S = {(Head, 1), (Head, 2), (Head, 3), (Head, 4), (Tail, 1), (Tail, 2), (Tail, 3), (Tail, 4)} (ii) n(S) = 8 Contoh CONTOH

Matematik Tingkatan 4 Bab 9 123 9.2 Peristiwa Bersandar dan Peristiwa Tak Bersandar Dependent Events and Independent Events Buku Teks m/s 246 – 252 1 Tentukan sama ada setiap pasangan peristiwa bergabung berikut ialah peristiwa bersandar atau peristiwa tak bersandar. SP: 9.2.1 TP2 Mudah Determine whether each of the following pairs of combined events is dependent event or independent event. (a) Mendapat dua guli biru apabila dua biji guli dikeluarkan secara rawak dari sebuah beg yang mengandungi lima guli merah dan enam guli biru satu demi satu dengan pemulangan. Obtaining two blue marbles when two marbles are drawn randomly from a bag containing five red marbles and six blue marbles, one by one with replacement. Peristiwa tak bersandar kerana kebarangkalian mendapat guli biru yang pertama tidak mempengaruhi kebarangkalian mendapat guli biru yang kedua. Independent event because the probability of obtaining the first blue marble does not affect the probability of obtaining the second blue marble. (b) Mendapat butang putih pada kali pertama dan butang kuning pada kali kedua apabila dua butang dikeluarkan dari sebuah kotak yang mengandungi empat butang putih dan lapan butang kuning tanpa pemulangan. Obtaining a white button for the first time and a yellow button for the second time when two buttons are drawn from a box containing four white buttons and eight yellow buttons without replacement. Peristiwa bersandar kerana kebarangkalian mendapat butang putih mempengaruhi kebarangkalian mendapat butang kuning. Dependent event because the probability of obtaining the white button affects the probability of obtaining the yellow button. (c) Mendapat dua keping kad yang berhuruf sama apabila dua keping kad dipilih secara rawak daripada kad berlabel dengan huruf “B, E, R, S, A, T, U” satu demi satu dengan pemulangan. Obtaining two cards with the same letters when two cards are chosen randomly from the cards labelled with letters “B, E, R, S, A, T, U” one after another with replacement. Peristiwa tak bersandar kerana kebarangkalian memilih kad yang pertama tidak mempengaruhi kebarangkalian memilih kad yang kedua. Independent event because the probability of choosing the first card does not affect the probability of choosing the second card. (d) Mendapat dua bola yang berwarna sama apabila dua biji bola dikeluarkan dari sebuah kotak yang mengandungi tiga bola hijau dan empat bola merah tanpa pemulangan. Obtaining two balls of the same colour when two balls are drawn from a box containing three green balls and four red balls without replacement. Peristiwa bersandar kerana kebarangkalian mendapat bola hijau yang pertama mempengaruhi kebarangkalian mendapat bola hijau yang kedua. Dependent event because the probability of obtaining the first green ball affects the probability of obtaining the second green ball. Mendapat angka dalam lambungan syiling adil dan nombor 5 dalam lambungan dadu adil. Obtaining head when tossing a fair coin and number 5 when tossing a fair dice. Peristiwa tak bersandar kerana kebarangkalian mendapat angka dalam lambungan syiling adil tidak mempengaruhi kebarangkalian mendapat nombor 5 dalam lambungan dadu adil. Independent event because the probability of obtaining head when tossing a fair coin does not affect the probability of obtaining number 5 when tossing a fair dice. Contoh Info Digital 9.2 CONTOH

Matematik Tingkatan 4 Bab 9 124 2 Tentu sahkan konjektur tentang rumus kebarangkalian peristiwa bergabung berikut. SP: 9.2.2 TP3 Sederhana Verify conjecture about the formula of probability of the following combined events. (a) Kotak D mengandungi lima kad berlabel dengan nombor “1, 2, 4, 5, 7” dan kotak E mengandungi lima kad berlabel dengan huruf “S, I, H, A, T”. Sekeping kad dipilih secara rawak dari kotak D dan E masing-masing. Tentu sahkan konjektur rumus kebarangkalian untuk mendapat nombor perdana dan huruf vokal dengan menyenaraikan semua kesudahan yang mungkin. Box D contains five cards labelled with numbers “1, 2, 4, 5, 7” and box E contains five cards labelled with letters “S, I, H, A, T”. A card is chosen randomly from boxes D and E respectively. Verify the conjecture about the formula of probability of getting a prime number and a vowel letter by listing all the possible outcomes. Kaedah I/ Method I: P(nombor perdana/ a prime number) = 3 5 P(huruf vokal/ a vowel letter) = 2 5 P(nombor perdana dan huruf vokal): P(a prime number and a vowel letter): 3 5 × 2 5 = 6 25 Kaedah II/ Method II: Kesudahan yang mungkin/ Possible outcomes: {(2, I), (2, A), (5, I), (5, A), (7, I), (7, A)} n(S) = 5 × 5 = 25 P(nombor perdana dan huruf vokal)= 6 25 P(a prime number and a vowel letter) ∴ Kedua-dua kaedah menghasilkan keputusan yang sama. Both methods produce the same result. (b) Sebuah kotak mengandungi empat helai tuala merah, dua helai tuala ungu dan tiga helai tuala krim. Shiha memilih dua helai tuala dari kotak tersebut secara rawak. Tuala pertama dipulangkan ke dalam kotak tersebut sebelum tuala kedua dipilih. Tentu sahkan konjektur rumus kebarangkalian untuk mendapat kedua-dua tuala berwarna krim dengan menyenaraikan semua kesudahan yang mungkin. A box contains four red towels, two purple towels and three cream towels. Shiha chooses two towels from the box randomly. The first towel is returned into the box before the second towel is chosen. Verify the conjecture about the formula of probability of getting both cream towels by listing all the possible outcomes. Kaedah I/ Method I: P(tuala krim pertama/ first cream towel) = 3 9 = 1 3 P(tuala krim kedua/ second cream towel) = 3 9 = 1 3 P(dua tuala krim/ two cream towels) = 1 3 × 1 3 = 1 9 Kaedah II/ Method II: Kesudahan yang mungkin/ Possible outcomes = {(K1 , K1 ), (K1 , K2 ), (K1 , K3 ), (K2 , K1 ), (K2 , K2 ), (K2 , K3 ), (K3 , K1 ), (K3 , K2 ), (K3 , K3 )} n(S) = 9 × 9 = 81 P(dua tuala krim/ two cream towels) = 9 81 = 1 9 ∴ Kedua-dua kaedah menghasilkan keputusan yang sama./ Both methods produce the same result. Kotak X mengandungi sembilan keping kad berlabel dengan huruf “C, E, M, E, R, L, A, N, G” dan kotak Y mengandungi lima kad berlabel dengan nombor “2, 5, 7, 8, 12”. Sekeping kad dipilih secara rawak dari kotak X dan Y masing-masing. Tentu sahkan konjektur rumus kebarangkalian untuk mendapat huruf “M” dan gandaan 2 dengan menyenaraikan semua kesudahan yang mungkin. Box X contains nine cards labelled with letters “C, E, M, E, R, L, A, N, G” and box Y contains five cards labelled with numbers “2, 5, 7, 8, 12”. A card is chosen at random from boxes X and Y respectively. Verify the conjecture about the formula of probability of getting letter “M” and a multiple of 2 by listing all the possible outcomes. Kaedah I/ Method I: P(huruf “M”/ letter “M”) = 1 9 P(gandaan 2/ a multiple of 2) = 3 5 P(huruf “M” dan gandaan 2) = 1 9 × 3 5 = 1 15 P(letter “M” and a multiple of 2) Kaedah II/ Method II: Kesudahan yang mungkin/ Possible outcomes: {(M, 2), (M, 8), (M, 12)} n(S) = 9 × 5 = 45 P(huruf “M” dan gandaan 2) = 3 45 = 1 15 P(letter “M” and a multiple of 2) ∴ Kedua-dua kaedah menghasilkan keputusan yang sama. Both methods produce the same result. Contoh CONTOH

Matematik Tingkatan 4 Bab 9 125 3 Selesaikan masalah yang berikut. SP: 9.2.3 TP3 TP4 Sederhana Solve the following problems. (a) Dua biji dadu adil telah dilambung secara serentak. Hitung kebarangkalian dadu pertama menunjukkan nombor 4 manakala dadu kedua menunjukkan nombor perdana. Two fair dice are tossed simultaneously. Calculate the probability that the first dice shows number 4 and the second dice shows a prime number. P(nombor 4)/ P(number 4) = 1 6 P(nombor perdana)/ P(a prime number) = 3 6 = 1 2 P(nombor 4 dan nombor perdana) = 1 6 × 1 2 P(number 4 and a prime number) = 1 12 (b) Di dalam sebuah almari terdapat tiga helai baju berwarna ungu, empat helai baju berwarna biru dan lima helai baju berwarna putih. Dua helai baju telah dipilih secara rawak dari almari tersebut, satu demi satu tanpa pemulangan. Hitung kebarangkalian bahawa kedua-dua helai baju tersebut adalah sama warna. A cupboard contains three purple shirts, four blue shirts and five white shirts. Two shirts are chosen from the cupboard one by one randomly without replacement. Calculate the probability that both shirts are of the same colour. P[(U, U) atau (B, B) atau (P, P)] P[(U, U) or (B, B) or (P, P)] = ( 3 12 × 2 11) + ( 4 12 × 3 11) + ( 5 12 × 4 11) = 6 + 12 + 20 132 = 38 132 = 19 66 Rajah berikut menunjukkan tiga keping kad berlabel dengan nombor dan empat keping kad berlabel dengan huruf masing-masing di dalam kotak R dan S. The following diagram shows three cards labelled with numbers and four cards labelled with letters in boxes R and S respectively. 4 1 9 Kotak R/ Box R U N T A Kotak S/ Box S Sekeping kad dipilih secara rawak masing-masing dari kotak R dan S. Hitung kebarangkalian mendapat A card is chosen at random from boxes R and S respectively. Calculate the probability of getting (a) kuasa dua sempurna dan huruf T, (b) nombor ganjil dan huruf konsonan. a perfect square and letter T, an odd number and a consonant letter. (a) P(kuasa dua sempurna/ a perfect square) = 1 P(huruf T/ letter T) = 1 4 P(kuasa dua sempurna dan huruf T) = 1 × 1 4 P(a perfect square and letter T) = 1 4 (b) P(nombor ganjil/ an odd number) = 2 3 P(huruf konsonan/ a consonant letter) = 2 4 = 1 2 P(nombor ganjil dan huruf konsonan) = 2 3 × 1 2 P(an odd number and a consonant letter) = 1 3 Kaedah Alternatif/ Alternative method: (a) Kuasa dua sempurna dan huruf T Perfect squares and letter T = {(1, T), (4, T), (9, T)} n(S) = 3 × 4 = 12 P(kuasa dua sempurna dan huruf T) = 3 12 P(a perfect square and letter T) = 1 4 (b) Nombor ganjil dan huruf konsonan Odd numbers and consonant letters = {(1, N), (1, T), (9, N), (9, T)} n(S) = 3 × 4 = 12 P(nombor ganjil dan huruf konsonan) = 4 12 P(an odd number and a consonant letter) = 1 3 Contoh CONTOH

Matematik Tingkatan 4 Bab 9 126 (c) Dua batang pen telah dipilih secara rawak dari sebuah kotak yang mengandungi 10 pen merah dan 15 pen biru. Hitung kebarangkalian mendapat Two pens are chosen randomly from a box that contains 10 red pens and 15 blue pens. Calculate the probability of getting (i) kedua-dua pen berwarna biru, both blue pens, (ii) sebatang pen biru dan sebatang pen merah. a blue pen and a red pen. (i) P(B, B) = 15 25 × 14 24 = 210 600 = 7 20 (ii) P(B, M) = 15 25 × 10 24 = 150 600 = 1 4 (d) Di dalam sebuah beg terdapat 25 biji guli putih dan 20 biji guli biru. Cindy mengambil dua biji guli secara rawak dari beg tersebut satu demi satu tanpa pemulangan. Hitung kebarangkalian dia mendapat A bag has 25 white marbles and 20 blue marbles. Cindy randomly takes two marbles from the bag one by one without replacement. Calculate the probability that she gets (i) sebiji guli putih dan sebiji guli biru, a white marble and a blue marble, (ii) dua biji guli putih. both white marbles. (i) P(P, B) = 25 45 × 20 44 = 500 1 980 = 25 99 (ii) P(P, P) = 25 45 × 24 44 = 600 1 980 = 10 33 (e) Rajah berikut menunjukkan kotak D yang mengandungi enam keping kad berlabel dengan huruf “B, E, R, S, I, H”. Dua keping kad dikeluarkan secara rawak dari kotak tersebut satu demi satu tanpa pemulangan. The following diagram shows box D containing six cards labelled with letters “B, E, R, S, I, H.” Two cards are drawn randomly from the box one after another without replacement. B E R S I H Kotak D/ Box D Hitung kebarangkalian mendapat kad pertama berhuruf vokal dan kad kedua berhuruf konsonan. Calculate the probability of getting the first card with vowel and the second card with consonant. P(vokal dan konsonan) = 2 6 × 4 5 P(vowel and consonant) = 4 15 (f) Kebarangkalian Jess bangun lewat ialah 0.3. Wakilkan peristiwa Jess bangun lewat bagi dua hari berturut-turut dengan menggunakan gambar rajah pokok. Kemudian, hitung kebarangkalian Jess bangun lewat dua hari berturut-turut. The probability for Jess to wake up late is 0.3. Use a tree diagram to represent the possible outcome that Jess will wake up late for two consecutive days. Then, calculate the probability that Jess will wake up late for two consecutive days. Hari pertama First day Hari kedua Second day Kesudahan Outcomes 0.3 0.3 0.3 0.7 0.7 0.7 L L (L, L) (L, L�) (L�, L) (L�, L�) L L� L' L� L = Bangun lewat/ Wake up late L� = Tidak bangun lewat/ Does not wake up late P(L, L) = 0.3 × 0.3 = 0.09 CONTOH

Matematik Tingkatan 4 Bab 9 127 (g) Rajah berikut menunjukkan empat keping kad berlabel dengan huruf dan tiga keping kad berlabel dengan nombor masing-masing di dalam kotak F dan G. The following diagram shows four cards labelled with letters and three cards labelled with numbers in boxes F and G respectively. C U B A Kotak F/ Box F 3 6 12 Kotak G/ Box G Sekeping kad dipilih secara rawak masing-masing dari kotak F dan G. Hitung kebarangkalian mendapat A card is chosen at random from boxes F and G respectively. Calculate the probability of getting (i) huruf vokal dan nombor genap, a vowel letter and an even number, (ii) huruf konsonan dan faktor bagi 9, a consonant letter and a factor of 9, (iii) huruf konsonan dan gandaan 6. a consonant letter and a multiple of 6. (i) P(huruf vokal dan nombor genap): P(a vowel letter and an even number): 2 4 × 2 3 = 1 3 (ii) P(huruf konsonan dan faktor bagi 9): P(a consonant letter and a factor of 9): 2 4 × 1 3 = 1 6 (iii) P(huruf konsonan dan gandaan 6): P(a consonant letter and a multiple of 6): 2 4 × 2 3 = 1 3 (h) Sebuah beg mengandungi 10 biji bola pingpong dengan dua warna yang berlainan. Diberi tiga biji bola pingpong adalah berwarna putih dan dua biji bola pingpong dikeluarkan satu demi satu tanpa pemulangan. Hitung kebarangkalian A bag contains 10 ping pong balls of two diffrent colours. Given three of them are white and two balls are drawn one after another without replacement. Calculate the probability that (i) bola pingpong pertama berwarna jingga dan bola pingpong kedua berwarna putih, the first ping pong ball is orange and the second ping pong ball is white, (ii) kedua-dua biji bola pingpong bukan berwarna putih. both ping pong balls are not white. Bola pertama First ball Bola kedua Second ball Kesudahan Outcomes P P (P, P) (P, J) (J, P) (J, J) P J J J 3 10 2 9 7 9 3 9 6 9 7 10 P = Putih/ White J = Jingga/ Orange (i) P(J, P) = 7 10 × 3 9 = 7 30 (ii) P(J, J) = 7 10 × 6 9 = 42 90 = 7 15 CONTOH

Matematik Tingkatan 4 Bab 9 128 (i) Terdapat 17 biji gelas di dalam sebuah kotak dan tiga daripada gelas itu telah pecah. Dua biji gelas telah dipilih secara rawak dari kotak tersebut. Hitung kebarangkalian There are 17 glasses in a box and three of them are broken. Two glasses are chosen at random from the box. Calculate the probability that (i) kedua-dua biji gelas telah pecah, both glasses are broken, (ii) tiada gelas yang pecah, none of the glasses are broken, (iii) kedua-dua biji gelas telah pecah atau satu daripada gelas itu telah pecah. both glasses are broken or one of the glasses is broken. (i) P(kedua-dua gelas pecah)/ P(both glasses are broken): 3 17 × 2 16 = 6 272 = 3 136 (ii) P(tiada gelas yang pecah)/ P(none of the glasses are broken): 14 17 × 13 16 = 182 272 = 91 136 (iii) P(kedua-dua gelas pecah atau satu daripada gelas pecah): P(both glasses are broken or one of the glasses is broken): ( 3 17 × 2 16) + ( 3 17 × 14 16) = 6 + 42 272 = 48 272 = 3 17 9.3 Peristiwa Saling Eksklusif dan Peristiwa Tidak Saling Eksklusif Mutually Exclusive Events and Non-Mutually Exclusive Events Buku Teks m/s 253 – 261 1 Tentukan sama ada pasangan peristiwa berikut ialah peristiwa saling eksklusif atau peristiwa tidak saling eksklusif. Determine whether the following pairs of events are mutually exclusive events or non-mutually exclusive events. SP: 9.3.1 TP2 Mudah Seorang atlet dipilih secara rawak daripada pasukan renang. An athlete is chosen randomly from a swimming team. • A ialah peristiwa atlet yang dipilih berasal dari Melaka. A is an event where the athlete chosen comes from Malacca. • B ialah peristiwa atlet yang dipilih seorang lelaki. B is an event where the athlete chosen is a man. • C ialah peristiwa atlet yang dipilih seorang wanita. C is an event where the athlete chosen is a woman. (i) A dan / and B Peristiwa tidak saling eksklusif kerana peristiwa A dan B boleh berlaku bersama. Non-mutually exclusive event because events A and B can occur together. (ii) A dan/ and C Peristiwa tidak saling eksklusif kerana peristiwa A dan C boleh berlaku bersama. Non-mutually exclusive event because events A and C can occur together. (iii) B dan/ and C Peristiwa saling eksklusif kerana peristiwa B dan C tidak boleh berlaku bersama. Mutually exclusive event because events B and C cannot occur together. Contoh Info Digital 9.3 CONTOH