20111101151326 final hbmt4403_jan2010_updated

CONFIDENTIAL / SULIT

JANUARY SEMESTER 2010 / SEMESTER JANUARI 2010

FINAL EXAMINATION / PEPERIKSAAN AKHIR

COURSE / KURSUS : TEACHING MATHEMATICS IN FORM SIX

CODE / KOD : HBMT4403

DATE / TARIKH :

DURATION / TEMPOH :

TIME / MASA :

INSTRUCTIONS TO CANDIDATES / ARAHAN KEPADA CALON

1. This question paper is set in English (in bold print) and Bahasa Melayu. ANSWER using ONE

LANGUAGE ONLY either ENGLISH or BAHASA MELAYU.

Kertas soalan ini disediakan dalam Bahasa Inggeris (dalam cetakan bold) dan Bahasa Melayu. JAWAB dalam SATU

BAHASA SAHAJA sama ada BAHASA INGGERIS atau BAHASA MELAYU.

2. This question paper consists of THREE Parts - PART A, PART B and PART C. Read

CAREFULLY the instructions for each PART.

Kertas soalan ini terdiri daripada TIGA Bahagian - BAHAGIAN A, BAHAGIAN B dan BAHAGIAN C. Baca

DENGAN TELITI arahan bagi setiap BAHAGIAN.

3. Write your answers in the Answer Booklet provided.

Tulis jawapan anda dalam Buku Jawapan yang dibekalkan.

This question paper consists of TWELVE PAGES of questions printed on both sides of the paper,

excluding this page.

Kertas soalan ini mengandungi DUA BELAS MUKA SURAT soalan yang dicetak pada kedua-dua belah muka surat, tidak

termasuk muka surat ini.

HBMT4403_SEM JAN10/F_AH

2…

PART A / BAHAGIAN A

INSTRUCTIONS / ARAHAN

Part A contains FIVE questions. Answer ALL questions.

Bahagian A mengandungi LIMA soalan. Jawab SEMUA soalan.

QUESTION/ SOALAN: Marks/ Markah

1. Using the first principles, show and explain that if ,uvy = then

dx

duv

dx

dvu

dx

dy+=

Menggunakan Prinsip Pertama, tunjuk dan terangkan jika diberi ,uvy =

maka dx

duv

dx

dvu

dx

dy+=

( 4 )

2. Given that xy cos= , show that 014

4

2

23 =++

y

dx

ydy

Diberi xy cos= , tunjukkan bahawa 014

4

2

23 =++

y

dx

ydy

( 4 )

3. With the help of a diagram, explain to your students the definition of

a parabola and state the general equation.

Dengan bantuan rajah, terangkan kepada pelajar anda definisi parabola

dan nyatakan persamaan am bagi parabola.

( 4 )


3…

4. Explain with the help of a suitable example the integration by parts.

Terangkan dengan menggunakan contoh yang sesuai kaedah pengamiran

bahagian demi bahagian.

( 4 )

5. Illustrate with an example each, the difference between permutation

and combination.

Tunjukkan dengan satu contoh untuk setiap kes, perbezaan di antara

permutasi dan kombinasi.

( 4 )

(Total/ Jumlah: 20)


4…

PART B / BAHAGIAN B


Part B contains FIVE questions. Answer THREE questions ONLY.

Bahagian B mengandungi LIMA soalan. Jawab TIGA soalan SAHAJA.


1. (a) You are teaching your class the theory of integration.

(i) Explain indefinite integration as the reverse process of

differentiation and give a simple example.

(ii) If ).( ,52

)(2

xfx

xxf ′

−= find Hence, evaluate

( )( )

dxx

xx∫−

−

−2

1 252

5

(b) Sketch on the same coordinate system, the curves

176 2 ++= xxy and 134 2 +−−= xxy . Calculate the area of the

region bounded by the two curves.

( 5 )

( 5 )

( 10 )

(Total/ Jumlah: 20)

(a) Anda sedang mengajar kelas anda teori pengamiran.

(i) Jelaskan pengamiran tak tentu sebagai proses berbalik bagi

pembezaan dan berikan satu contoh yang ringkas.

(ii) Jika ).( cari,52

)(2

xfx

xxf ′

−= Kemudian, dapatkan nilai

( )( )

dxx

xx∫−

−

−2

1 252

5.

(b) Lakarkan lengkung 176 2 ++= xxy dan 134 2 +−−= xxy pada

satu sistem koordinat yang sama. Kira luas kawasan yang dibatasi

oleh dua lengkung tersebut.


5…

2. (a) Explain to your students the difference between exclusive and

non-exclusive sets. Use Venn diagrams to illustrate the

difference.

(b) Show to your students how to solve this problem step-by-step

using probability tree:

In a small town where Ameer stays, the rain falls randomly three days

in a week (*a week = 7 days). Of the total number of days that rain,

80% of the days Ameer takes bus to school. If it does not rain, Ameer

cycles to school. For every 5 days Ameer cycles to school, 4 days he

saves the pocket money.

For one particular day, find the probability that:

(i) Ameer does not take bus to school.

(ii) Ameer saves his pocket money because cycling to school.

( 6 )

( 14 )

(Total/ Jumlah: 20)

(a) Terangkan kepada pelajar anda perbezaan di antara set-set yang

eksklusif dan tidak eksklusif. Tunjukkan perbezaan tersebut dengan

menggunakan gambar rajah Venn.

(b) Tunjukkan kepada pelajar anda bagaimana menyelesaikan masalah

di bawah langkah demi langkah menggunakan gambar rajah pokok

kebarangkalian:

Di pekan kecil tempat Ameer tinggal, hujan turun secara rawak tiga hari

dalam seminggu (*seminggu = 7 hari). Daripada keseluruhan jumlah hari

yang hujan, 80% daripadanya Ameer pergi ke sekolah dengan menaiki

bas. Jika tidak hujan, Ameer akan ke sekolah menaiki basikal. Bagi setiap

5 hari Ameer menaiki basikal ke sekolah, dia dapat menyimpan duit poket

selama 4 hari.


6…

Cari kebarangkalian bagi satu hari tertentu bahawa:

(i) Ameer tidak menaiki bas ke sekolah

(ii) Ameer menyimpan duit poket kerana menaiki basikal ke

sekolah.

3. You are planning to do a classroom project with your students. The

project consists of seven activities. The duration of each activity is

shown in the table below. You would like to apply the network model

for this project.

Activity Preceding activities Duration (days)

A

B

C

D

E

F

G

-

-

-

B

A,D

C

E,F

2

5

1

10

3

6

8

(a) Show and explain how to draw a network diagram for the

project and to determine the minimum completion time of the

project.

(b) Calculate the total float for each activity and state the critical

path of the project. Then, explain the importance of “float” for

project management.

( 10 )

( 10 )

(Total/ Jumlah: 20)


7…

Anda bercadang untuk menjalankan satu projek bersama-sama pelajar di

dalam sebuah kelas. Projek tersebut mengandungi tujuh aktiviti. Tempoh

masa untuk setiap aktiviti ditunjukkan di dalam jadual di bawah. Anda

ingin mengaplikasikan model jaringan untuk projek berkenaan.

Aktiviti Aktiviti terdahulu Tempoh (hari)

A

B

C

D

E

F

G

-

-

-

B

A,D

C

E,F

2

5

1

10

3

6

8

(a) Tunjukkan dan terangkan gambar rajah jaringan bagi projek di atas

dan bagaimana untuk menentukan masa penyelesaian terpendek

bagi projek tersebut.

(b) Kirakan jumlah apungan bagi setiap aktiviti dan nyatakan laluan

kritikal bagi projek tersebut. Kemudian terangkan kepentingan

apungan dalam pengurusan projek.

4. How would you guide your students to solve the following problems?

(a) Box A contains 3 cards numbered 1,2 and 3 ; Box B contains 4

cards numbered 1,2,3, and 4. The cards are mixed thoroughly in

each box, and a card is then selected at random from each box.

Let X denotes the sum of numbers obtained in each trial.

i) Tabulate the distribution of X. Show that it is a

probability distribution, that is X is a discrete random

variable.

ii) Find )5( >XP

( 7 )

( 3 )


8…

(b) The discrete random variable X has a probability function given

by ( )

=

=

otherwise

xifx

xf

0

5,4,3,2,115

Find mean, variance and standard deviation of X.

( 10 )

(Total/ Jumlah: 20)

Bagaimana anda membimbing pelajar anda menyelesaikan masalah-

masalah berikut?

(a) Kotak A mengandungi 3 kad bernombor 1, 2 dan 3 ; kotak B

mengandungi kad bernombor 1, 2, 3 dan 4. Kad-kad tersebut

digoncang bebas dalam setiap kotak masing-masing, kemudian satu

kad dikeluarkan secara rawak daripada setiap kotak.

Jadikan X mewakili jumlah nombor yang diperolehi dalam setiap

cubaan.

i) Bina jadual taburan X. Tunjukkan bahawa ia adalah taburan

kebarangkalian, di mana X ialah pembolehubah rawak

diskrit.

ii) Cari )5( >XP

(b) Suatu pembolehubah rawak diskrit X mempunyai fungsi

kebarangkalian berikut:

( )

=

=

selainnya

xjikax

xf

0

5,4,3,2,115

Cari min, varian dan sisihan piawai bagi X.


9…

5. You are showing in the class the steps taken to solve the following

problems. Explain in detail.

(a) In a mathematics examination, 55% of the students do not pass.

If 20 students takes the examination, find the probability that:

i) Exactly 15 students pass

ii) at least two students pass

iii) more than 16 students pass

(b) Given that ).16,100(~ NX Use the standardised normal table to

find:

(i) )112108( << XP

(ii) ( )4100 <−XP

(10 )

(10 )

(Total/ Jumlah: 20)

Anda sedang menunjukkan di dalam kelas anda langkah-langkah

menyelesaikan masalah-masalah berikut. Terangkan dengan terperinci.

(a) Dalam satu ujian Matematik, 55% daripada pelajar yang menduduki

ujian tersebut tidak lulus. Jika satu kelas yang terdiri daripada 20

orang pelajar mengambil ujian tersebut, cari kebarangkalian

bahawa:

i) 15 orang pelajar lulus

ii) Sekurang-kurangnya dua orang pelajar lulus

iii) Lebih daripada 16 orang pelajar lulus.

(b) Diberi ).16,100(~ NX Dengan menggunakan jadual taburan

normal piawai, cari:

(i) )112108( << XP

(ii) ( )4100 <−XP


10…

PART C / BAHAGIAN C


Part C contains TWO questions. Answer ONE question ONLY.

Bahagian C mengandungi DUA soalan. Jawab SATU soalan SAHAJA.


1. Linear programming deals with problems of maximization and

minimization in real life situation. As a teacher, you should be able to

explain and transform the problems into mathematical model as well

as creating a simple problem (question) using a model.

Referring to the mathematical model given below, answer the

following questions.

Maximise 21 53 xxZ +=

Subject to:

0,

1823

122

4

21

21

2

1

≥

≤+

≤

≤

xx

xx

x

x

(a) Create a simple real life problem / question to explain how the

above model is developed.

(b) Transform the mathematical model to standard form and

explain how to solve the linear programming problem using

simplex method.

( 4 )

(16)

(Total/ Jumlah: 20)


11…

Pengaturcaraan linear berkait dengan penyelesaian masalah memaksima

dan meminima dalam situasi kehidupan sebenar. Sebagai guru, anda

sepatutnya mampu menerangkan dan menukar masalah tersebut kepada

bentuk model matematik dan juga membina soalan ringkas daripada model

matematik sedia ada.

Merujuk kepada model matematik yang diberikan di bawah, jawab soalan-

soalan berikut:

Maximise 21 53 xxZ +=

Subject to:

0,

1823

122

4

21

21

2

1

≥

≤+

≤

≤

xx

xx

x

x

(a) Bina satu soalan / masalah ringkas dalam situasi kehidupan sebenar

untuk menjelaskan bagaimana model di atas diperolehi.

(b) Tukarkan model matematik yang diberikan kepada bentuk piawai

dan terangkan bagaimana untuk menyelesaikan masalah

pengaturcaraan linear tersebut dengan menggunakan kaedah

simpleks.


12…

2. Linear regression attempts to model the relationship between two

variables by fitting a linear equation to observed data. One variable is

considered to be an independent (explanatory) variable, and the other

is considered to be a dependent variable. The most common method

for fitting a regression line is the method of least-squares.

Based on the information above and the data given below, explain to

your students and show in detail how to answer the following

questions:

X 1 2 3 4 4 6

Y 62 78 70 90 93 103

a) Find the least square regression line equation of y on x for the

above data.

b) Explain the meaning of correlation coefficient ( r ) and coefficient

of determination ( r2 ). Then, find and interpret the values for r

and r2 of the linear regression equation above.

( 12 )

( 8 )

(Total/ Jumlah: 20)

Regresi linear adalah kaedah untuk memodelkan hubungan di antara dua

pembolehubah dengan memadankan suatu garis linear kepada data yang

diperolehi. Satu pembolehubah disebut sebagai pembolehubah tak

bersandar dan satu lagi disebut pembolehubah bersandar. Kaedah yang

paling kerap digunakan untuk memadankan garis regresi linear ialah

kaedah kuasa dua terkecil.

Berdasarkan maklumat di atas dan set data yang diberikan di bawah,

jelaskan kepada pelajar anda dan tunjukkan dengan terperinci cara-cara

bagaimana untuk menjawab soalan-soalan berikut:

X 1 2 3 4 4 6

Y 62 78 70 90 93 103

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(a) Dapatkan persamaan garis regresi kuasa dua terkecil y melawan x

bagi data di atas.

(b) Terangkan apakah yang dimaksudkan dengan pekali korelasi, r dan

pekali penentu, r2. Kemudian dapatkan dan jelaskan nilai r dan r

2

bagi persamaan garis regresi linear di atas.

END OF QUESTION PAPER / KERTAS SOALAN TAMAT

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KURSUS: HBMT 4403 TEACHING MATHEMATICS IN FORM SIX

MARKING SCHEME FINAL EXAMINATION

SEMESTER: JANUARY 2010

Note: Students are not allowed to get this marking scheme.

PART A

INSTRUCTIONS

Part A contains FIVE questions. Answer ALL questions.

QUESTION 1 (Chapter 1)

From the first principles, we know that ( )

x

xfxxfxf

x δδ

δ

)(lim)(

0

−+=′

→

Given that y=uv,

x

xvxuxxvxxuuv

dx

d

x δδδ

δ

)()()()(lim)(

0

−++=∴

→

to change this fraction into an equivalent one that contains difference quotients for the

derivatives of u and v, we subtract and add )()( xvxxu δ+ in the numerator:

( ) ( )

( ) ( )x

xuxxuxv

x

xvxxvhxu

x

xuxxuxv

x

xvxxvxxu

x

xvxuxvxxuxvxxuxxvxxuuv

dx

d

xxx

x

x

δδ

δδ

δδ

δδ

δ

δδδδδ

δδδ

δ

δ

)(lim)(

)()(limlim

)()()(

)(lim

)()()()()()()()(lim)(

000

0

0

−+•+

−+•+=

−++

−++=

−+++−++=∴

→→→

→

→

as xδ approaching zero, )( xxu δ+ approaches )(xu because u , being differentiable at x , is

continuous at .x The two fractions approach the value of dxdv / at x and dxdu / at .x

∴dx

duv

dx

dvuuv

dx

d+=)(

(proving steps : 3 marks + 1 mark explanation)

(4 marks)

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( )

( )

014

1

cos1

cos

cos1

4

1cos44

cos2

sincos2

2

1

cos

sincos2

1sincoscos

2

1

cos

sin

2

1

sincos2

1

cos

4

2

23

4

2

2

3

2

2

3

2

23

2

3

22

2

1

2

1

2

1

2

1

2

1

=++

∴

−−=

−−=

+

−

=

∴

+

−=

−−

−=∴

−=

−=∴

=

−

−

ydx

ydy

y

x

x

xx

dx

ydy

x

xx

x

xxxxx

dx

dy

x

x

xxdx

dy

xy

(4 marks)

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(diagram : 1 mark)

A parabola is the locus points which move in such a way that its distance from a fiexd point

(focus) is always equal to its perpendicular distance from a fixed straight line (directrix) not

containing the focus.

the fixed point Q(a,0) is called the focus

the fixed line x = -a is called the directrix

0 is called the vertex. (explanation : 2 marks)

General equation of a parabola axy 42 = (equation :1 mark)

(4 marks)

y

x

P(x,y)

Q(a,0)

R

x=-a

0

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QUESTION 4 (Chapter 3 )

Using the formula ∫∫ ′−=′ dxxfxgxgxfdxxgxf )().()().()().(

one part is to be differentiate and the other part is to be integrated.

Example :

Integrate ( )dxx∫ +1ln 2 . Let

( )xxgdxx

xxf

dxxgandxxf

=+

=′

=′+=

)(21

1)(

)(1ln)(

2

2

( ) ( )

( )( ) cxxxx

dxx

xx

dxx

xxxxdxx

++−+=+

−−+=

+

−+=+∴

−

∫

∫∫

12

2

2

2

22

tan221ln

1

1121ln

1

21ln1ln

Correct formula + explanation = 1 mark

Suitable example = 3 marks

(4 marks)


A permutation is an arrangement of a group of objects in a particular order.

Example of permutation problem: How many number can be formed using digit 1, 3,

and 8?

A combination is a selection of a group of objects where the order of objects selected is

immaterial.

Example of combination problem: How many different committees of three can be

formed by selecting the members from 6 persons?

Explanations = 2 marks

Suitable examples = 2 marks

(4 marks)

[Total: 20 marks]

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BAHAGIAN B

INSTRUCTION

Part B contains FIVE questions. Answer THREE questions ONLY.


(a)

(i) If we differentiate )(xf , we will write )()(

xfdx

xdf ′= , which is called the derivative

of )(xf . Conversely, in integration, )(xf is the anti derivative of )(xf ′ and the process

of finding ∫ dxxf )( for a given function )(xf is called integration.

If )(xfdx

dy=

Then ∫ ∫ ∫=⇒= dxxfythereforedxxfdy )(,)( CxF += )(

Example :

If ( ) 23 34 xxdx

d=+ . Then dxx∫ 23 = Cx ++ )4( 3

(Explanation = 3 marks)

(Example : 2 mark)

(ii)

( )( )( )

( )( )

( )22

2

2

2

2

52

52

52

102

52

)2(522)(

52)(

−

−=

−

−=

−

−−=′

−=

x

xx

x

xx

x

xxxxf

x

xxf

-1

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( )( )

( )

( )

6

11

3

1

2

1)4(

2

1

522

1

52

)5(

522

1

52

52

2

1

52

)5(

2

1

22

1 2

2

22

−=

+

−−

+−=

+

−=

−

−∴

+

−=

−

−=

−

−∴

∫∫

∫

∫∫

CC

Cx

xdx

x

xx

Cx

x

x

xxdx

x

xx

(5 marks)

(b)

12

7 when

24

25 -of valueminimum thehas curve This

24

25

12

76

176

2

2

−=⇒−

+=

++=

xx

xxy

8

3 when

16

25 of valuemaximum thehas curve This

16

25

8

34

134

2

2

−=⇒+

+−=

+−−=

xx

xxy

Solve the simultaneous equation to find the intersection points of the two curves:

( )( )2134

1176

2

2

+−−=

++=

xxy

xxy

( )10

0110

01010

134176

2

22

−==∴

=+∴

=+∴

+−−=++=∴

xorx

xx

xx

xxxxy

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(Procedure + graph = 5 marks)

Area of the region bounded by the two curves:

( ) ( )

2

0

1

3

0

1

2

20

1

2

3

55

3

10

53

10

1010

176134

units

xx

dxxx

dxxxxx

=

+−=

−−=

−−=

++−+−−=

−

−

−

∫

∫

(5 marks)

(Total: 20 marks)

1

-1

1762 ++= xxy

134 2 +−−= xxy

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(a) Set A and Set B are exclusive sets if their intersection is empty, thus φ=∩ BA .

If Set A and Set B are not exclusive, then Set A and Set B have some elements

in common. The following Venn Diagrams show the difference of exclusive sets

and not exclusive sets.

(3 marks)

Exclusive Sets Not exclusive sets

(1.5 marks each diagram = 3 marks)

(b) Let

moneypocket his savenot doesAmeer event the:

moneypocket his savesAmeer event the:

school tocyclesAmeer event the:

school tobus sAmeer takeevent the:

rainnot doesit event that the:

rainsit event that the:

S

S

C

B

R

R

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The probability tree diagram :

`

(6 marks)

a)

( ) ( )

657.035

23

7

6.4

7

4

7

6.0

0.17

42.0

7

3

school) tobus not take does(

atau

BRPBRP

P

=

=

+=

×+×=

∩+∩=

(4 marks)

7

3

R

R

7

4

0.1C

8.0

2.0

B

C

S

S

S

S

5

4

5

4

5

1

5

1

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0

A

D

B

C F

G

E

3

b)

( ) ( )

5257.0350

184

35

4.18

35

16

35

4.2

5

40.1

7

4

5

42.0

7

3

money)pocket his savesAmeer (

atau

SCRPSCRP

P

=

=

+=

××+××=

∩∩+∩∩=

(4 marks)

(Total: 20 marks)


(a)

Minimum completion time = 26 days

(Correct diagram = 4 marks)

(Calculation = 6 marks)

1

0 0

2

5 5

4

15 15

3

12 1

5

18 18

6

26 26

3

8 5

10 2

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(b)

Activity Total float

A

B

C

D

E

F

G

13

0

9

0

0

19

0

Critical path : B-D-E-G

(4 marks)

Floats are useful for management because they indicate how much an activity can be

delayed without delaying the whole project. Resources can be redeployed from an

activity with float to be used elsewhere, possibly to ensure that a critical activity finishes

on time.

(6 marks)

(Total: 20 marks)


(a) (i)

(4 marks)

So the distribution of X is:

X=x 2 3 4 5 6 7

P(X=x)

12

1

12

2

12

3

12

3

12

2

12

1

X=x Outcomes corresponding to

X=x

P(X=x)

2

3

4

5

6

7

(1,1)

(1,2), (2,1)

(1,3), (3,1), (2,2)

(2,3), (3,2), (1,4)

(3,3), (2,4)

(3,4)

1/12

2/12

3/12

3/12

2/12

1/12

SCH BSMG

ACCP 1015 AU12003A - HR

This is a probability distribution, that X is a discrete random variable, because P(X=x) ≥ 0

for every value of x and P(S)=1

(4 marks)

(ii) P(X>5) = P(X=6) + P(X=7)

= 4

1

12

1

12

2=+ (2 marks)

(c)

∑ ==== )()( xXxPXEXMean µ

3

11

15

55

15

44

15

33

15

22

15

11 =

+

+

+

+

=

(4 marks)

9

14

3

1115

1515

525

15

416

15

39

15

24

15

11)(

)()(

2

2

222

=

−=∴

=

+

+

+

+

=

−==

Varians

XE

XEXVar µσ

(4 marks)

3

14

9

14)(,, ofdeviation Standard ==XSDX σ (2 mark)

(Total: 20 marks)

SCH BSMG

ACCP 1015 AU12003A - HR


(5

markah)

(a) Let q to be the proportion of students fail, th

(i)

−<<

−=<<

16

100112

16

100108)112108( ZPXP

)32( <<= ZP

02145.0

00135.00228.0

)3()2(

=

−=

−= QQ

(5 marks)

(ii) ( ) ( )410044100 <−<−=<− XPXP

( )10496 <<= XP

−<<

−=

16

100104

16

10096ZP

( ) ( )

( ) ( )

( )

( )[ ][ ]

( )

( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

0003.0

45.055.045.0)20(55.045.0)190(55.045.0)1140(

45.055.045.055.045.055.045.0

)20()19()18()17(16)

9999.0

0001.01

)55.0)(45.0)(20()55.0(1

)1()0(1

11)2()

0049.0

55.045.0!5!5

!20

55.045.0)15()

)45.0,20(~ then pass, whostudents ofnumber theis If

45.055.01

55.0

20119218317

20

20

20119

19

20218

18

20317

17

20

1920

515

515

15

20

=

+++=

+++=

=+=+=+==>

=

−=

+−=

=+=−=

≤−=≥

=

=

==

=−=

=

CCCC

XPXPXPXPXPiii

XPXP

XPXPii

CXPi

BXX

pSo

q

marks 3

marks 2

SCH BSMG

ACCP 1015 AU12003A - HR

6826.0

)1587.0(21

)1(21

)11(

=

−=

−=

<<−=

Q

ZP

(5 marks)

(Total: 20 marks)

SCH BSMG

ACCP 1015 AU12003A - HR

PART C

INSTRUCTIONS

Part C contains TWO questions. Answer ONE question ONLY.


(a) Accept any SUITABLE and CORRECT answer.

(4 marks)

(b) Change to standard form:

053 21 =−− xxZ

411 =+ Sx

122 22 =+ Sx

1823 321 =++ Sxx

Simplex tableau:

iteration

Basic

variable

Coefficient of Right

Side Z

1x 2x 1S 2S 3S

0

Z

1S

2S

3S

1

0

0

0

-3

1

0

3

-5

0

2

2

0

1

0

0

0

0

1

0

0

0

0

1

0

4

12

18

iteration

Basic variable

Coefficient of Right Side

Z 1x 2x 1S 2S 3S

1

Z

1S

2x

3S

1

0

0

0

-3

1

0

3

0

0

1

0

0

1

0

0

5/2

0

1/2

-1

0

0

0

1

30

4

6

6

SCH BSMG

ACCP 1015 AU12003A - HR

The table is optimal since there are no more negative values entries in the z row.

Optimum solution : 1x =2 ; 2x =6 and profit (Z) to be maximized is = RM36,000.

(workout & answer : 16 marks)

iteration

Basic

variable

Coefficient of Right

Side Z

1x 2x 1S 2S 3S

2

Z

1S

2x

1x

1

0

0

0

0

0

0

1

0

0

1

0

0

1

0

0

3/2

1/3

1/2

-1/3

1

1/3

0

1/3

36

2

6

2

(Total: 20 marks)


(a)

y = ax+b as a regression line:

x y x² y² xy

1 62 1 3844 62

2 78 4 6084 156

3 70 9 4900 210

4 90 16 8100 360

4 93 16 8649 372

6 103 36 10609 618

∑x = 20 ∑y = 496 ∑x²=82 ∑y²= 42186 ∑xy = 1778

( )

33.15

6

2082

2

2

2

=

−=

−=∑ ∑n

xxSxx

SCH BSMG

ACCP 1015 AU12003A - HR

( )

33.1183

6

49642186

2

2

2

=

−=

−=∑ ∑n

yySyy

( )( )

66.124

6

)496)(20(1778

=

−=

−=∑ ∑∑n

yxxySxy

59.55

)33.3(13.867.82

13.8

33.15

66.124

=

−=

−=

=

==

xayb

Sxx

Sxya

→ regression line equation : y= 55.59+8.13x

(workout and answer : 12 marks)

(b)

The correlation coefficient, r, is the strength of the relationship between the two variables in

the regression equation and is always a value between -1 and 1, inclusive. The regression

coefficient is the slope of the line of the regression equation.

(2 marks)

The coefficient of determination, r2, measures how good the estimated regression equation is.

The higher the r2, the more confidence one can have in the equation. Statistically, the

coefficient of determination represents the proportion of the total variation in the y variable

that is explained by the regression equation. It has the range of values between 0 and 1

(2 marks)

correlation coefficient , r = 9256.033.118333.15

66.124=

×=

SxxSyy

Sxy

Indicates a strong positif linear relationship between X and Y

SCH BSMG

ACCP 1015 AU12003A - HR

(2 marks)

(c) coefficient of determination , r²

8567.0

33.1183

33.15/66.124/ 22

=

=

=

Syy

SxxSxy

Indicates that 85.67% of the total variation is explained by the regression line.

(2 marks)

(Total: 20 marks)

SCH BSMG

ACCP 1015 AU12003A - HR

COURSE: HBMT4403 TEACHING MATHEMATICS IN FORM SIX

1. DIFFERENTIATION FORMULA

2. INTEGRATION FORMULA

Basic functions

=

Basic Functions

Addition

Linearity

Product Rule

Quotient Rule

Chain Rule

Inverse

Trig Functions

SCH BSMG

ACCP 1015 AU12003A - HR

Trigonometric functions

3. DISCRETE PROBABILITY DISTRIBUTION

∑ ==== )()()( xXxPXEXMean µ

222 )()( µσ −== XEXVar

2)(deviation Standard σ=X

SCH BSMG

ACCP 1015 AU12003A - HR

4. REGSRESSION ANALYSIS

( )∑ ∑−=

n

xxSxx

2

2 ; ( )

∑ ∑−=n

yySyy

2

2 ; ( )( )

∑ ∑∑−=n

yxxySxy

xaybandSyy

Sxxawherebaxy −==+= ;

SxxSyy

Sxyr =

=

Syy

SxxSxyr

/2

2

20111101151326 final hbmt4403_jan2010_updated

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