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BAHAN PROGRAM INTERVENSI

PPSMI 2007 UNTUK MURID

TINGKATAN LIMA

ADDITIONAL MATHEMATICS


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MINIMUM SYLLABUS REQUIREMENT

1.FUNCTIONS:

Express the relation between the following pairs of sets in the form of arrow diagram,ordered pair and graph.

Arrow diagram Ordered pair Graph

a ) Set A = {

Kelantan, Perak ,

Selangor }

Set B = { Shah Alam ,Kota Bharu ,Ipoh }

Relation: City of thestate in Malaysia

b )Set A ={triangle,rectangle,

pentagon }

Set B = { 3,4,5 }

Relation : Number of

Sides

Determine domain , codomain , object, image and range of relation.


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1. Diagram 1 shows the relation between set P and set Q.

a. Domain = }

b. Codomain = { } c. Object =

d. Image =

e. Range =...f. Ordered Pairs =

Classifying the types of relations

3

2

-2

5

4

3

9

Set P Set QDiagram 1

-3 1

1


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State the type of the following relations

a)

..

b )

.

c)

..

d)

2.QUADRATIC EQUATIONS:

A. TO EXPRESS A GIVEN QUADRATIC EQUATION IN GENERAL FORM

x

4

162

2

x

364

6

32 4

9

4

9

Prime

-3-3

x

3

24

-29

-3

Even

x X

x X2Type of number


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ax2 + bx + c = 0

Example 1

.x2 = 5x 9

.x2 5x + 9 = 0

Compare with the general form

ax2 + bx c = 0

Thus, a = 1, b = -5 and c = 9

Example 2

4x =x

xx 22

4x(x) = x2 2x

4x2 - x2 2x = 0

3x2 2x = 0Compare with the general form

Thus, a = 3, b = - 2 and c = 0

Exercises

Express the following equation in general form and state the values of a, b and c


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1. 3x =x2

5

.2. (2x + 5) =x

7

3. x( x + 4 ) = 3 .4. (x 1)(x + 2) = 3

5.x

4=

x

x

+

5

3 6. x2 + px = 2x - 6

7. px (2 x) = x 4m 8. (2x 1)(x + 4) = k(x 1) + 3

9. (7 2x + 3x2) =

3

1+x10. 7x 1 =

x

xx 22


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B. FORMING QUADRATIC EQUATIONS FROM GIVEN ROOTS

Example 1

3 , 2

x = 3 , x = 2

x - 3 = 0 , x-2 = 0(x-3)(x-2)=0

x2 5x + 6 = 0

Example 2

1, - 3

x = 1 , x = -3

x 1 = 0 , x + 3 = 0

(x 1 ) ( x + 3 ) = 0x2 + 2x 3 = 0

a) 4 , -7

b)

2 ,3

1

c)

3

1,

2

1

d)

5

1,

3

2

e) 4 , 0


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3. QUADRATIC FUNCTIONS: INEQUALITIES

Example

Find the range of values ofx for which 01522 > xx

Solution

01522

> xxLet ( ) 1522 = xxxf

= ( )( )53 + xxWhen ( ) 0=xf( )( ) 053 =+ xx

3=x or 5

For 01522 > xx

5>x or 3


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4.SIMULTANEOUS EQUATIONS: EXPRESS ONE UNKNOWN IN TERM OF

THE OTHER UNKNOWN

Guidance Example

1 Arrange the linear equation such thatone of the two unknowns becomes the

subject of the equation.

(avoid fraction if possible)

x + 2y = 1

x =

2 Substitute the new equation from step 1

into the non-linear equation . Simplifyand express in the form

ax2 + bx + c = 0.

( )2 + 4y2 = 13

= 0

3 Solve the quadratic equation by

factorisation, completing the square orby using the formula

(2y 3)( ) = 0,

y =

2

3or

4 Substitute the values of the unknown

obtained in step 3 into the linearequation.

When y =2

3,

x = 1 2( ) =

When y = ,

x =


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5.INDICES AND LOGARITHM: INDICES

Examples Exercises

Solve each of the following equations

1. 33x = 8133x = 34

3x = 4

x =34

1. 9x = 271-x

2. 2x . 4x+1 = 642x . 22 (x+1) = 26

x + 2x + 2 = 6

3x = 4

x =3

4

2. 4x . 8x -1 = 4

3. 0168 1 = +xx

( ) ( ) 022 143 = +xx

( ) ( )143

22+

=xx

44322 +=

xx

3x = 4x + 4

x = - 4

3. 5x - 25x+1 = 0


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6.COORDINATE GEOMETRY

Examples Solution

1. Determine whether the straight lines2y x = 5 and x 2y = 3 are parallel.

2y x = 5,

y = 52

1+x ,

2

11 =m

x 2y = 3

y = 3

2

1x ,

2

12 =m

Since 21 mm = , therefore the straight lines 2y x = 5

and x 2y = 3 are parallel.

2. Given that the straight lines 4x + py = 5and 2x 5y 6 = 0 are parallel, find the

value of p.

Step1: Determine the gradients of both straightlines.

4x + py = 5

y =p

xp

54 + ,p

m4

1 =

2x 5y 6 = 0y = 3

2

5+x ,

2

52 =m

Step 2: Compare the gradient of both straight lines.

Given both straight lines are parallel, hence

21mm =

5

24 =p

p = -10

3. Find the equation of the straight linewhich passes through the point P(-3, 6)and is parallel to the straight line

4x 2y + 1 = 0.

4x 2y + 1 = 0, y = 2x +2

1.

Thus, the gradient of the line, m = 2.

Therefore, the equation of the line passing throughP(-3, 6) and parallel to the line 4x 2y + 1 = 0 is

y - 6 = 2 (x - -3)

y = 2x + 12.


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Examples Solution

1. Determine whether the straight lines

3y x 2 = 0 and y + 3x + 4 = 0 areperpendicular.

3y x 2 = 0

y =3

2

3

1+x ,

3

11 =m

y + 3x + 4 = 0

y= 3x 4, 32 =m

)3(3

121 =mm = -1.

Hence, both straight lines are perpendicular.

Examples Solution

2. Find the equation of the straight line

which is perpendicular to the straight

line x + 2y 6 = 0 and passes through

the point (3, -4).

x + 2y 6 = 0

y = 32

1+ x ,

2

11 =m

Let the gradient of the straight line which is

perpendicular = 2m

221 m

= -1

2m =

The equation of the straight line

=

y =


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7.STATISTICS

The number of vehicles that pass by a toll plaza from 1 p.m to 2 p.m. for 60consecutive days is shown in the table below.

Number of vehicles Number of

days5950 46960 107970 248980 169990 6

Calculate the median of the number of cars using formula.Solution :

Number of

vehicles

Number of days

(f)

Cumulative

frequency5950 4 46960 10 (14)7970 (24) 388980 16 ( )9990 6 ( )

Step 1 : Median class is given by = 302

60

2

TTTn ==

Therefore, the median class is 7970

Step 2 : Median = cf

Fn

Lm

+ 2

= (___)

+ 24

142

60

( __ )

= 76.17

L = lower boundary of the median

class = 69.5

n =

F = cumulative frequency before the

median class = 14

fm

= frequency of the median class

=24c = size of the median class

= upper boundary lower

boundary

= 79.5 69.5

= 10

Median lies in this

interval


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To estimate the mode using a histogram

Modal class = 7970

(c)

Class boundary Number of days(frequency)

49.5 59.5 4

10

24

16

6


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(c) The histogram is shown below

49.5 59.5 69.5 79.5 89.5 99.5 Number of vehicles

Mode = 76

Frequency

5

10

15

20

25


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8.CIRCULAR MEASURE

Convert Measurements in radians to degrees and vice versa.

Convert the following angles in radians to degrees and minutes.

a. 1.5 rad b. 0.63 rad

c. rad2

d. rad2

3

Convert the following angles to radians.

a. 500 b. 124.30

c. 72035 d. 285021


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Arc Length of a circleFind the length of arc.

1.

0.5 rad

8 cm

Q

P

O

2.

152

6.4 cmO

BA

Complete the table below by finding the values of , r or s.

r s

1. 1.5 rad 9 cm

2. 14 cm 30 cm

3. 2.333 rad 35 cm


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Complete the table below, given the areas and the radii of the sectors andangles subtended.

2

2

1rA= , is in radians

Area of sector Radius Angle subtended

1. 38.12 cm 500

2. 90 cm2 9.15 cm

3. 72 cm2 =1.64 rad

4. 18cm2 6.5 cm

5. 200 cm2

1.778 rad

6. 145 cm2 8 cm


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9.DIFFERENTIATION:

1. y = 10

dxdy =

2. y = 5x

dxdy =

3. f(x) = -2 3x

f (x)=4. y =

x

7

dx

dy=

5.3

31)(x

xf =

f (x)=

6. xxy +=2

4

dx

dy=

7. =

+ x

xx

dx

d5

12

2

2

=

=

dx

dy

xxy )23(

9. Given xxy 43 2 = , find the value of

dx

dywhenx =2.

10. Given ( )21)( xxxf += , find the valueof ).1('and)0(' ff

11.INDEX NUMBER

Always change

a fractional

function to the

negative index

before finding

differentiation

8.


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The table shows the price of 3 types of goods: A, B and C in the year 2005 and 2006.

Types of good Price Price index in 2006

(Base year = 2005)2005 2006

A RM 1.20 RM 1.60 z

B x RM 2.30 110

C RM 0.60 y 102

Find the value of x, y and z

Calculate the composite index for each of the following data

Index number, I 120 110 105

Weightage, W 3 4 3

1.PROGRESSIONS

1. Find the 9th term of the arithmetic

progression.

2. Find the 11th term of the arithmetic

progression.


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2, 5 , 8 , ..

Solution:a = 2

d = 5-2=3

9 2 (9 1)3T = +

= _______

5

3, ,2, ........2

3. For the arithmetic progression

0.7, 2.1 , 3.5, .. ,find the 5th term .

4. Find the thn term of the arithmetic

progression

1

4,6 ,9, .....2

5. Find the 7 th term of the geometric

progression.

- 8, 4 , -2 , ..

Solution:

a = - 8 r =8

4

=

2

1

T7 = (-8)(2

1

)7-1

=8

1


progression.

16, -8, 4,

7. For the geometric progression

9

4,3

2, 1 , .. ,find the 9 th term .


progression50, 40, 32.

Find the sum to infinity of geometric progressions

1

aS

r =

sum to infinity

a = first term

r = common ratio

S =


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Find the sum to infinity of a givengeomertric progression below:

Example:

2 26, 2, , , .......

3 9

a = 6

2 1

6 3r

= =

1

6=

11- -

3

9=

2

aSr

=

1. 24, 3.6, 0.54, .

2. 81, -27,9, ..

3.1 1 1

, , ,.......2 4 8

..

* example on recurring decimals

2.LINEAR LAW

STEPS TO PLOT A STRAIGHT LINE

Using a graphpaper.


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QUESTION

SOLUTION

x 2 3 4 5 6

y 2 9 20 35 54

The above table shows the experimental values of two variables, xand y. It is know that x and y are related by the equation

y = px2 + qx

a) Draw the line of best fit forx

yagainst x

a) From your graph, find,i) p

ii) q

Table

Identify Y and X from part (a)

Construct a table

Follow the scale given.

Label both axes

Line of best fit

Determine : gradient m

Y-intercept c

Non- linear


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STEP 1

y = px2 + qx

xy =

xpx

2

+xqx

x

y= px + q

Y = mX + c

Note : For teachers reference

STEP 2

x 2 3 4 5 6

y 2 9 20 35 54

x

y

1 3 5 7 9

STEP 3

Reduce the non-linearTo the linear form

The equation is divided throughout by x

To create a constant that is free from xOn the right-hand side i.e, q

Linear form

Y = mX + c

construct table

Using graph paper,

- Choose a suitable scale so that the graph

drawn is as big as possible.

- Label both axis

-Plot the graph of Y against X and drawthe line of best fit


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x

y

12

10

x

8

x

6

x

4

x

2

x

2 3 4 5 6x

- 2

- 4

1


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3.INTEGRATION

STEP 4

Gradient , p =26

19

= 2

y- intercept = q= -3

From the graph,

find p and q

Construct a right-angled triangle,So that two vertices are on the line

of best fit, calculate the gradient, p

Determine the y-intercept, q

from the straight line graph


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1. Given that2

1( ) 3f x dx = and

2

3( ) 7f x dx = . Find

(a) [ ]2

1the value of k if ( ) 8kx f x dx =

(b) [ ]3

15 ( ) 1f x dx

Answer : (a) k =22

3(b) 48

2. Given that4

0( ) 3f x dx = and

4

0( ) 5g x dx = . Find

(a)

4 0

0 4( ) ( )f x dx g x dx (b) [ ]

4

03 ( ) ( )f x g x dx

Answer: (a) 15

(b) 4

4.VECTOR


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VECTOR IN THE CARTESIAN COORDINATES

1. State the following vector in terms in~i and

~

j and also in Cartesian coordinates

Example Solutions

~

22

0OA i

= =

~

03

3OB j

= =

~ ~

3 4

3

4

OP p i j

= = +

= Exercise Solutions

(a) OP

= (b)OQ

=

(c)OR

=(d)

OS

=

(e)OT

= (f)OW

=

~

j5

4

3

2

1

543210

~

p

0

B

P

A

1

4

3

2

1

2

SR

P

Q

-1-3 -2 -1

T

W

31 4

~i

~

j

-2

O


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2. Find the magnitude for each of the vectors

Example

3 ~ ~2i j =

2 23 2

13 unit

+

=

(a) ~ ~2 5i j+ =

(b) ~ ~5 12i j = (c) ~i j =

3. Find the magnitude and unit vector for each of the following

Example

~ ~ ~

3 4r i j= +

Solution :

2 2

~

~ ~ ~

Magnitude, 3 4

= 5

1unit vector, r, (4 3 )

5

r

i j

= +

= +

(a) ~ ~ ~2 6r i j=

(b)~

6

3a

=

(c)

~

1

2h

=

SPM 2003/no. 12 / paper 1.


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1. Diagram 2 shows two vectors, and QOOP .

Express

(a) OP in the formx

y

,

(b) OQ in the formxi +yj. [ 2 marks]

5.TRIGONOMETRIC FUNCTIONS

P(5, 3)

y

Q(-8, 4)

xO


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To sketch the graph of sine or cosine function , students are encouraged to follow the

steps below.

1. Determine the angle to be labeled on x-axis.

eg : Function angle

y = sin x x = 90o

y = cos 2x 2x = 90o

x = 45o

y = sin x2

3x

2

3= 90o

x = 60o

2. Calculate the values of y for each value of x by using calculator

eg : y = 1 2 cos 2x

x 0 45 90 135 180 225 270 315 360

y -1 1 3 1 -1 1 3 1 -1

3. Plot the coordinates and sketch the graph

6.PERMUTATIONS AND COMBINATIONS

45 90 135 180 225 270 315 360 x

y

3

2

1


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1. The number of ways of arranging all the

alphabets in the given word.

Example Solution:

6! = 6.5.4.3.2.1

= 720

2. The number of ways of arranging four

of the alphabets in the given word so that

last alphabet is S

Example Solution:

The way to arrange alphabet S = 1

The way to arrange another 3 alphabets= 5

P 3

The number of arrangement = 1 x 5 P 3 = 60

3. How many ways to choose 5 books

from 20 different books

Example solution:

The number of ways= 20 C 5

= 15504

4. In how many ways can committee of 3

men and 3 women be chosen from a group

of 7 men and 6 women ?

Example Solution:

The numbers of ways = 7 C 3 x6 C 3

= 700

7.PROBABILITY


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Alternative method

Black

Yellow

Black

Yellow

Black

Yellow

10

6

10

4

10

4

10

4

10

6

10

6

Question Answer

The above figure shows sixnumbered cards. A card is

chosen at random. Calculatethe probability that thenumber on the chosen card

(a) is a multiple of 3 and

a factor of 12

(b) is a multiple of 3 or afactor of 12.

Let

A represent the event that the number on the chosen card

is a multiple of 3, andB represent the event that the number on the chosen card

is a factor of 12.A = {3, 6, 9}, n(A)= 3B = {2, 3, 4, 6}, n(B) = 4

A B = {3, 6}A B = {2, 3, 4, 6, 9}

(a) P(A B) =3

1

6

2 = .

(b) P(A B) =6

5

P(A B) = P(A) + P(B) P(A B)

=6

2

6

4

6

3 +

=6

5.

Question Solution

A box contains 5 red balls, 3 yellow balls

and 4 green balls. A ball is chosen atrandom from the box. Calculate the

probability that the balls drawn neither ayellow nor a green.

P (yellow) =3

12.

P(green) =

4

12

P(yellow or green) =3

12+

4

12=

7

12.

No Questions Solutions

1.

Box C contains 4 black marbles and 6

2 3 4 6 8 9

10

4


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yellow marbles. A marbles is chosen

at random from box C, its colour isnoted and the marbles is noted and the

marbles is returned to the box. Then a

second marbles is chosen. Determine

the probability that(a) both the marbles are black.

(b) the two balls are of differentcolours.

(c) at least one of the balls chosen is

yellow.

(a) P(black black)=10

4

10

4 =25

4

(b) P(same colours)

= P(black black) + P(yellow yellow)

=

25

4+

10

6

10

6=

25

13.

(c) 1 P(both blacks) = 1 25

4=

25

21

8.PROBABILITY DISTRIBUTIONS

Example 1 :


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Find the value of each of the following probabilities by reading the standardised normal

distribution table.

(a) P(Z > 0.934)

(b) P(Z 1.25)

Solution

(b) P(Z 1.25) = 1 P(Z > 1.25)= 1 0.1057

= 0.8944

(c) P(Z - 0.23)

1.251.25


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Solution

(c) P(Z - 0.23) = 1 P(Z < - 0.23)= 1 P(Z > 0.23)

= 1 0.40905

= 0.59095

(d) P(Z > - 1.512)

Solution

(d) P(Z < - 1.512) = P(Z > 1.512)

= 0.06527

(e) P(0.4 < Z < 1.2)

Solution

(e) P(0.4 < Z < 1.2) = P(Z > 0.4) P(Z > 1.2)

= 0.3446 0.1151

= 0.2295

(f) P(- 0.828 < Z - 0. 555)

Solution

-1.512 1.512

-0.230.23

0.4 1.2

0.4 1.2


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(f) P(- 0.828 < Z - 0. 555) = P(Z > 0.555) P(Z > 0.828)= 0.28945 0.20384

= 0.08561

(g) P(- 0.255 Z < 0.13)

Solution

(g) P(- 0.255 Z < 0.13) = 1 P(Z < - 0.255) P(Z > 0.13)= 1 P(Z > 0.255) P(Z > 0.13)

= 1 0.39936 0.44828

= 0.15236

14.3 Score- z

Example 2 :

-0.828 -0.555 0.555 0.828

-0.255 0.13 0.13-0.255


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Find the value of each of the following :

(a) P(Z z) = 0.2546(b) P(Z < z) = 0.0329

(c) P(Z < z) = 0.6623

(d) P(z < Z < z 0.548) = 0.4723

Solution

(a) P(Z z) = 0.2546Score-z = 0.66

(b) P(Z < z) = 0.0329Score-z = -1.84

(c) P(Z < z) = 0.66231 - P(Z > z) = 0.6623

P(Z > z) = 1 0.6623

= 0.3377

Score-z = 0.419

(d) P(z < Z < z 0.548) = 0.4723

1 P(Z < z) P(Z > 0.548) = 0.4723

1 P(Z < z) 0.2919 = 0.4723P(Z < z) = 1 0.2919 0.4723

= 0.2358

Score-z = -0.72

Normal DistributionType 1

P( Z > positive no)

P ( Z > 1.2 ) = 0.1151

Type 6

P (Negative no < Z < Negative no )

Type 1

P ( Z > K ) = less than 0.5

P ( Z > K ) = 0.2743

z

0.2546


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.....................................................

Type 2

P(Z < negative no)

P ( Z < - 0.8 ) = P (Z > 0.8)

= 0.2119

.....................................................

Type 3

P ( Z < positive no)

P ( Z < 1.3 )

= 1 P ( Z>1.3)

= 1 0.0968

= 0.9032

.....................................................

.

Type 4.

P( Z > negative no)

P ( Z > - 1.4 )

= 1 P ( Z < -1.4 )

= 1 0.0808

= 0.9192

....................................................

Type 5

P( positive no < Z < positive

no)

P ( 1 < Z < 2 )

= P ( Z > 1 ) P ( Z > 2 )= 0.1587 0. 0228

= 0.1359

P ( -1.5 < Z < - 0.8 )

= P ( 0.8 < Z < 1.5 )

= P ( Z > 0.8 ) P ( Z > 1.5 )

= 0.2119 0.0668 = 0.1451

.....................................................

.

Type 7

P ( negative no < Z < postive no )

P ( -1.2 < Z < 0.8 )

= 1 P ( Z > 0.8) P ( z < -1.2 )

= 1 P ( Z > 0.8 ) P ( Z >

1.2 )

= 1 0.2119 0.1151

=0.673

K = 0.6

......................................................

Type 2

P ( Z < K ) = less than 0.5

P( Z < K ) = 0.3446

P ( Z > - K ) = 0.3446

- K = 0.4

K = - 0.4

.......................................................

Type 3

P( Z < K ) = more than 0.5

P ( Z < K ) = 0.8849

P ( Z > K ) = 1 0.8849

= 0.1151

K = 1.2

......................................................

Type 4

P ( Z > K ) = more than 0.5

P ( Z > K ) = 0.7580

P( Z < K ) = 1 0.7580 = 0.2420

P ( Z > -k ) = 0.2420

- K= 0.7

K = - 0.7

10.LINEAR PROGRAMMING

Problem interpretation and the formation of the relevant equations or inequalities


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The table below shows the mathematical expressions for the different inequalities

used.

Mathematical Expressions Inequality

a y greater than xxy

>b y less than x xy

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