Post on 14-Apr-2018
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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
6. Find the 8 th term of the geometric
progression.
16, -8, 4,
7. For the geometric progression
9
4,3
2, 1 , .. ,find the 9 th term .
8. Find the 3 th term of the geometric
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