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5a)
No of red balls = 3
No of green balls = 5
No of blue balls = x
Prob.(red ball) = no of total outcome/no of possible outcome
Pr(red) = 3/3+5+x = 1/6
3/8+x = 1/6
6(3) = 1(8+x)
18 = 8 + x
X = 18 – 8 = 10
Therefore the no of blue ball = 10
(5b)
Probability of picking a green ball
P(g) = no of green balls/no of possible outcome
P(g) = 5/3+5+10 = 5/18
=5/18
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6ai)
F α M1M2/d²
F = KM1M2/d²
Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m
20 = k(25)(10)/5²
250k = 500
k = 500/250 = 2
Expression is
F = 2M1M2/d²
(6aii)
Making d subject
d = √2M1M2/F
d = √2 ×7.5×4/30
d = √60/30 = √2
d = √2m or 1.41m
(6b)
Draw the diagram
X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)
5x + 200 = 540
5x = 540 – 200
5x = 340
X = 340/5
X = 68
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8a)
1/3x – 1/4(x+2)>_ 3x -1⅓
1/3x – 1/4(x+2)>_3x – 4/3
Multiply through by the L. C. M(12), we have
4x – 3(x + 2)>_36x – 16
4x – 3x – 6 >_ 36x – 16
-6+16 >_36x + 3x – 4x
10 >_ 35x
35x _< 10 X = 10/35 X = 2/7 (8bi) Draw the triangle |AB|/66 = sin35 |AB| = 66sin35 = 66×0.5736 = 37.8576 Draw the right angled triangle |AD|/|AB| = Tan52 |AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m Height of tower = 48.45m ========================
(11ai) ar² = 1/4 ......(1) ar^5= 1/32 .....(2) Divide eqn (2) by eqn(1) ar^5/ar² = 1/32÷1/4 r³ = 1/32 × 4/1 r³= 1/8 r³ = 2-³ r = 2-¹ r = 1/2 Common ratio = 1/2 Put this into eqn (1) a(1/2)² = 1/4 a(1/4) = 1/4 a = (1/4)/(1/4) = 1 First term, a = 1
(11aii) Seventh term, T7 = ar^6 =(1)(1/2)^6 =1/64 (1b) Given : X = 2 and X = -3 (X - 2)(X + 3) = 0 X² + 3x - 2x - 6 , 0 X² + x - 6 = 0 Comparing with ax²+bx+c = 0 a = 1 b = 1 C = -6