WAEC SSCE - Further Mathematics - 2022

Vraag 1 Verslag

Express $ \frac{4\pi}{2} $ radians in degrees.

288º

200º

144º

120º

Antwoorddetails

$\frac{4 π}{2}$ = $\frac{4}{5} * 180$

= 144º

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Vraag 2 Verslag

The gradient of a function at any point (x,y) 2x - 6. If the function passes through (1,2), find the function.

x

^{2}

- 6x - 5

x

^{2}

- 6x + 5

x

^{2}

- 6x - 3

x

^{2}

- 6x + 7

Antwoorddetails

dy/dx = 2x - 6

y = ∫ 2x - 6

y = $\frac{2 x^{2}}{2} - 6 + c$ y = x $^{2}$ - 6x + c
passes through (1,2)
2 = 1 $^{2}$ - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x $^{2}$ - 6x + c

y = x $^{2}$ - 6x + 7

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Vraag 3 Verslag

The first, second and third terms of an exponential sequence (G.P) are (x - 4), (x + 2), and (3x + 1) respectively. Find the values of x.

\frac{- 1}{2}, 8

\frac{1}{2}, - 8

\frac{- 1}{2}, - 8

\frac{1}{2}, 8

Antwoorddetails

U1 = x - 4
U2 = x + 2
U3 = 3x + 1

$\frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}}$

$\frac{x + 2}{x - 4} = \frac{3 x + 1}{x + 2}$

(x+2)(x+2) = (x-4)(3x+1)
x $^{2}$ + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x $^{2}$ - 15x - 8 =0
2x $^{2}$ + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0

x = ( $\frac{- 1}{2}, 8$ )

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Vraag 4 Verslag

Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6

76

^{\circ}

53

^{\circ}

37

^{\circ}

14

^{\circ}

Antwoorddetails

tanθ

=

m1 - m21 + m1m2

y = 2x + 5
m1 = 2
2y = x - 6

y

=

12

x

-

3

m2

=

12

tanθ

=

2 -

\frac{1}{2}

1+2(

\frac{1}{2}

)

tanθ = $\frac{3}{2}$ ÷ (1+1)

tanθ = $\frac{3}{2}$ ÷ 2

tanθ

=

34

θ = $t a n^{- 1} (\frac{3}{4})$

θ = 36.87º
θ = 37º

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Vraag 5 Verslag

\[ \left(\frac{3\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{54}}{3\sqrt{5}}\right)^{-1} \]

\frac{5 \sqrt 3}{6}

\frac{3 \sqrt 15}{6}

\frac{5 \sqrt 6}{12}

\frac{5 \sqrt 3}{12}

Antwoorddetails

( $\frac{3 \sqrt 6}{\sqrt 5} + \frac{\sqrt 54}{3 \sqrt 5}$ ) $^{- 1}$

= $\frac{\sqrt 5 (3 \sqrt 5)}{3 \sqrt 6 + 3 \sqrt 6}$

= $\frac{3 * 5}{6 \sqrt 6} = \frac{5}{2 \sqrt 6}$

= $\frac{5 * 2 \sqrt 6}{2 \sqrt 6 + 2 \sqrt 6} = \frac{10 \sqrt 6}{4 * 6}$

= $\frac{5 \sqrt 6}{12}$

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Vraag 6 Verslag

Solve: $3^{2x-2}-28(3^{x-2})+3=0$

x = -2 or x = 1

x = 0 or x = -3

x = 2 or x = 1

x = 0 or x = 3

Antwoorddetails

$3^{2 x - 2} - 28 (3^{x - 2}) + 3 = 0$

$\frac{3^{2 x}}{3^{2}} - \frac{{28.3}^{x}}{3^{2}} + 3 = 0$

$\frac{3^{2 x}}{9} - \frac{{28.3}^{x}}{9} + 3 = 0$

let p = 3 $^{x}$

$\frac{p^{2}}{9} - \frac{28 p}{9} + 3 = 0$

multiply through by 9
p $^{2}$ - 28p + 27 = 0
p $^{2}$ - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3 $^{x}$
3 $^{x}$ = 1
3 $^{x}$ = 3 $^{0}$
x = 0
when p = 27
3x = 27
3x = 33
x = 3

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Vraag 7 Verslag

Simplify $ \frac{9 \ast 3^{n+1}-3^{n+2}}{3^{n+1}-3^n} $

3

9

27

81

Antwoorddetails

$\frac{9 * 3^{n + 1} - 3^{n + 2}}{3^{n + 1} - 3^{n}}$

= $\frac{3^{n} * 3^{n} * 3^{1} * - 3^{2} * 3^{2}}{3^{n} * 3^{1} - 3^{n}}$

= $\frac{3^{n} (3^{2} * 3^{1})}{3^{n} (3^{1} - 1)}$

= $\frac{27 - 9}{3 - 1}$

= $\frac{18}{2}$

= 9

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Vraag 8 Verslag

Given $\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix}\begin{vmatrix} -6 \\ k \end{vmatrix}\begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15$ Solve for k.

-8

-5

-4

-3

Antwoorddetails

$| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15$

$| \begin{matrix} 2 [- 6] & - 3 k \\ 1 [- 6] & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

$| \begin{matrix} - 12 & - 3 k \\ - 6 & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

-12 - 3k = 3

-3k = 3 + 12

k = $\frac{15}{- 3}$

k = -5

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Vraag 9 Verslag

A linear transformation T is defined by T: (x,y) → (3x - y, x + 4y). Find the image of (2, -1) under T.

(7, -2)

(5, -2)

(-2, 7)

(-7, 2)

Matrices And Linear Transformation

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Vraag 10 Verslag

Differentiate $ \frac{5x^3+x^2}{x} $, x ≠ 0 with respect to x.

10x + 1

10x + 2

x(15x + 1)

x(15x + 2)

Antwoorddetails

$\frac{5 x^{3} + x^{2}}{x}$ → $\frac{5 x^{3}}{x} + \frac{x^{2}}{x}$

5x $^{2}$ + x

Then dy/dx = 10x + 1

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Vraag 11 Verslag

If $U_n = kn^2 + pn$, $U_1 = -1$, $U_5 = 15$, find the values of k and p.

k = -1, p = 2

k = -1, p = -2

k = 1, p = -2

k = 1, p = 2

Antwoorddetails

Un = kn $^{2}$ + pn,

U $_{1}$ = -1,

U $_{5}$ = 15,

when n = 1
U1 = k(1) $^{2}$ + p(1) = -1
k + p = -1 --------eqn1
when n = 5
U $_{5}$ = k(5) $^{2}$ + p(5) = 15
25k + 5p = 15 --------eqn2
multiply eqn1 by 5 and eqn2 by 1
5k + 5p = -5 -------eqn3
25k + 5p = 15 -------eqn4
eqn4 - eqn3
20k = 20
k = 1
sub for k in eqn1
1 + p = -1
p = -1 -1 = -2

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Polynomial Functions

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Vraag 12 Verslag

Solve: $4\sin^2\theta + 1 = 2$, where $0º < \theta < 180º$

60º 0r 120º

30º 0r 150º

30º 0r 120º

60º 0r 150º

Antwoorddetails

4sin $^{2}$ θ + 1 = 2

4sin $^{2}$ θ = 2 - 1

4sin $^{2}$ θ = 1

$\sqrt{s} i n^{2} θ$ = $\sqrt{\frac{1}{4}}$

sinθ = $\frac{1}{2}$

θ = $s i n^{- 1} \frac{1}{2}$

θ = 30º 0r 150º

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Vraag 13 Verslag

If →PQ = -2i + 5j and →RQ = -i - 7j, find →PR

-3i + 12j

-3i - 12j

-i + 12j

i - 12j

Vectors

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Vraag 14 Verslag

Find the range of values of x for which $2x^2 + 7x - 15 \geq 0$.

x ≤ -5 or x ≥

\frac{3}{2}

x ≥ -5 or x ≤

\frac{3}{2}

-5 ≤ x ≤

\frac{3}{5}

\frac{3}{5}

≤ x ≤ -5

Antwoorddetails

2x $^{2}$ + 7x - 15 ≥ 0

2x $^{2}$ -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0

x ≤ -5 or x ≥ $\frac{3}{2}$

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Vraag 15 Verslag

The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.

4√3

2√6

3√2

2√3

Surds

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Vraag 16 Verslag

Evaluate $\int_{0}^{1} x^{2}(x^{3}+2)^{3}$

$\frac{56}{12}$

$\frac{65}{12}$

12

65

Antwoorddetails

$1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$ dx

let $u = x^{3} + 2, d u = 3 x^{2} d x$

when x = 1, u = 3

when x = 0, u = 2

dx = $\frac{d u}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{x^{2} [u]^{3}}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{u^{3}}{3}$ du

= $\frac{u^{4}}{3 * 4}$ $_{2}$ 3

$\frac{1}{12} [u^{4}]$ $_{2}$ 3

$\frac{1}{12} [3^{4} - 2^{4}]$

$\frac{1}{12} [81 - 16]$

$\frac{65}{12}$

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Integration

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Vraag 17 Verslag

Evaluate $4p_2 + 4C_2 - 4p_3$

18

6

-6

-18

Antwoorddetails

$4 p_{2} + 4 C_{2} - 4 p_{3}$

$n p_{r} = \frac{n!}{[n - r]!} a n d n C_{r} = \frac{n!}{[n - r]! r!}$

= $\frac{4!}{[4 - 2]!} + \frac{4!}{[4 - 2]! 2!} - \frac{4!}{[4 - 3]!} = \frac{4!}{2!} + \frac{4!}{2! 2!} - \frac{4!}{1!}$

= $\frac{4 * 3 * 2!}{2!} + \frac{4 * 3 * 2!}{2! 2!} - \frac{4 * 3 * 2 * 1}{1!}$

12 + 6 - 24 = -6

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Sets

Permutation And Combinations

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Vraag 18 Verslag

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

All strong wrestlers are weightlifters

Some strong wrestlers are not weightlifters

Some weak wrestlers are weightlifters

All weightlifters are wrestlers

Logical Reasoning

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Vraag 19 Verslag

Find the coefficient of $x^3y^2$ in the binomial expansion of $(x-2y)^5$

-80

10

40

80

Antwoorddetails

x $^{3}$ y $^{2}$ in (x-2y) $^{5}$

n = 5, r = 3, p = x, q = -2y

5C $_{3}$ * x $^{3}$ $^{2}$

5C $_{3}$ = $\frac{5!}{[5 - 3]! 3!}$

$\frac{5 * 4 * 3!}{2! 3!}$ → $\frac{5 * 4}{2}$

5C $_{3}$ = 10

: 5C $_{3}$ * x $^{3}$ -2y $^{2}$ = 10 * x $^{3}$ 4y $^{2}$

40x $^{3}$ y $^{2}$
the coefficient is 40

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Binomial Theorem

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Vraag 20 Verslag

A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that $d = t^2 + 3t$. Find the magnitude of F at time t.

0N

2N

3(2t + 3)N

6N

Antwoorddetails

F = m * a

d = t $^{2}$ + 3t.

a = $\frac{d^{2} d}{d t^{2}}$

$\frac{d [d]}{d t}$ = 2t + 3

$\frac{d^{2} d}{d t^{2}}$ = 2m/s $^{2}$

a = 2m/s $^{2}$

F = m * a

F = 3 × 2 = 6N

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Vraag 21 Verslag

Given that P = {x: x is a multiple of 5}, Q = {x: x is a multiple of 3} and R = {x: x is an odd number} are subsets of μ = {x: 20 ≤ x ≤ 35}, (P⋃Q)∩R.

{20, 21, 25, 30, 33}

{21, 25, 27, 33, 35}

{20, 21, 25, 27, 33, 35}

{21, 25, 27, 30, 33, 35}

Sets

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Vraag 22 Verslag

Which of the following is the semi-interquartile range of a distribution?

Mode - Median

Highest score - Lowest score

1/2 (Upper Quartile - Median)

1/2 (Upper Quartile - Lower Quartile)

Statistics

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Vraag 23 Verslag

Evaluate $ \int_{-1}^{0} (x + 1)(x - 2)\,dx $

7/6

5/6

-5/6

-7/6

Antwoorddetails

$\int_{- 1}^{0}$ (x + 1)(x - 2) dx

= $\int_{- 1}^{0}$ $x^{2} - x - 2$ dx

Integrated $x^{2} - x - 2$ = $\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2$

=

(0

-

0

-

0)

-

(

-13

-

12

+

2)

=

0

-

(

76

)

=

-76

Bekijk antwoord

Vraag 24 Verslag

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

\frac{1}{2}

m/s

\frac{1}{3}

m/s

2m/s

3m/s

Antwoorddetails

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v
12 = 24v
v = $\frac{1}{2}$ m/s

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Matrices And Linear Transformation

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Vraag 25 Verslag

If 36, $ \frac{9}{4} $ and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

9/16

81/16

9

\frac{9}{16}

Antwoorddetails

GP : 36, P, $\frac{q}{4}$ , q, ... p + q = ?

Recall,

common

ratio,

r

=

TnTn-1

=

T2T1

=

T3T2

=

T4T3

∴

P36

=

94

÷

p

;

p

^{2}

=

94

x

36

;

p

^{2}

=

81

p

=

9

∴

r

=

T2T1

=

936

=

14

Also

r

=

T4T3

=

q

÷

94

∴ $\frac{1}{4}$ = q ÷ $\frac{9}{4}$ ;

$\frac{9}{4}$ = 4q

16q

=

9

,

q

=

916

∴

p

+

q

=

9

+

916

=

9

916

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Vraag 26 Verslag

If $x^2+y^2-2x-6y+5=0$, evaluate dy/dx when x=3 and y=2.

2

-2

-4

4

Antwoorddetails

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

When differentiated:

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

where x=3 and y=2

2[3] + 2[2] - 8 = 0

6 + 4 - 8 = 2

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Vraag 27 Verslag

If $f(x-1)=x^3+3x^2+4x-5$, find $f(2)$

61

25

20

13

Antwoorddetails

x - 1 = 2
x = 3
f(2) = (3) $^{3}$ + 3(3) $^{2}$ + 4(3) - 5
f(2) = 27 + 27 + 12 - 5

= 61

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Vraag 28 Verslag

If $ \log_{10}(3x+1) + \log_{10}4 = \log_{10}(9x+2) $, find the value of x

\frac{1}{3}

1

2

3

Antwoorddetails

Explanation

$l o g_{10} (3 x + 1) + l o g_{10} 4 = l o g_{10} (9 x + 2)$

$l o g_{10} 4 (3 x + 1) = l o g_{10} (9 x + 2)$

4(3x+1) = 9x + 2

12x -4 = 9x + 2

12x - 9x = 2 + 4

3x = 6

x = 2

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Vraag 29 Verslag

Given that $ \frac{8x+m}{x^2-3x-4} \equiv \frac{5}{x+1}+\frac{3}{x-4} $

23

17

-17

18

Antwoorddetails

$\frac{8 x + m}{x^{2} - 3 x - 4} \equiv \frac{5}{x + 1} + \frac{3}{x - 4}$

$\frac{8 x + m}{x^{2} - 3 x - 4}$ ≡ $\frac{5 (x - 1) + 3 (x + 4)}{x^{2} - 3 x - 4}$

multiplying both sides by x2-3x-4
8x+m ≡ 5(x-4)+3(x+1)
8x + m ≡ 5x - 20 + 3x + 3
8x - 5x - 3x + m = -20 + 3
m = -17

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Polynomial Functions

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Vraag 30 Verslag

A particle moving with a velocity of 5m/s accelerates at $2\text{m/s}^2$. Find the distance it covers in 4 seconds.

16m

26m

36m

46m

Antwoorddetails

from the equation of motion

u = 5m/s, a = 2m/s $^{2}$ , t = 4s

s = ut + $\frac{1}{2} a t^{2}$

s = 5*4 + $\frac{1}{2} 2 * 4^{2}$

s = 20 + 16

s = 36m

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Vraag 31 Verslag

If $ g(x) = \sqrt{1-x^2} $, find the domain of $ g(x) $

x < -1 or x > 1

x ≤ -1 or x ≥1

-1 ≤ x ≤ 1

-1 < x < 1

Antwoorddetails

1 - x $^{2}$ ≥ 0
-x $^{2}$ ≥ -1
x $^{2}$ ≤ 1
√x $^{2}$ ≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1

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Surds

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Vraag 32 Verslag

The functions $f:x \to 2x^2 + 3x -7$ and $g:x \to 5x^2 + 7x - 6$ are defined on the set of real numbers, R. Find the values of $x$ for which $3f(x) = g(x)$.

x = -3 or -5

x = -3 or 5

x = 3 or -5

x = 3 or 5

Functions

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Vraag 33 Verslag

The equation of a circle is given as $2x^2 + 2y^2 - x - 3y - 41 = 0$. Find the coordinates of its centre.

(

\frac{- 1}{4}

,

\frac{3}{4}

)

(

\frac{1}{4}

,

\frac{3}{4}

(

\frac{- 1}{2}

,

\frac{3}{2}

)

. (

\frac{- 1}{2}

,

\frac{- 3}{2}

)

Antwoorddetails

2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

standard equation of circle
(x-a) $^{2}$ + (x-b) $^{2}$ = r $^{2}$
General form of equation of a circle.
x $^{2}$ + y $^{2}$ + 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

= x $^{2}$ + y $^{2}$ + 2gx + 2fy + c
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41 = 0
divide through by 2

g = $\frac{- 1}{4}$ ; 2g = $\frac{- 1}{2}$

f = $\frac{- 3}{4}$ ; 2f = $\frac{- 3}{2}$

a = -g → - $\frac{- 1}{4}$ ; = $\frac{1}{4}$

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is ( $\frac{1}{4}$ , $\frac{3}{4}$ )

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Vraag 34 Verslag

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

If a hunter is selected at random, find the probability that the hunter covered at least 6km.

\frac{3}{5}

\frac{2}{5}

\frac{3}{8}

\frac{9}{40}

Probability

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Vraag 35 Verslag

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.

1.63m

1.54m

1.52m

1.42m

Statistics

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Vraag 36 Verslag

Find the coefficient of $x^2$ in the binomial expansion of $\left(x+\frac{2}{x^2}\right)^5$

10

40

32

80

Antwoorddetails

$(x + \frac{2}{x^{2}})^{5}$

n = 5, r = 4, p = x and q = $\frac{2}{x^{2}}$

5C $_{4}$ x $^{4}$ ( $\frac{2}{x^{2}}$ )1 = 5C $_{4}$ $\frac{2 x^{4}}{x^{2}}$

5C $_{4}$ 2x $^{2}$ = $\frac{5!}{[5 - 4]! 4!}$ * 2x $^{2}$

$\frac{5 * 4!}{4!} * 2 x^{2}$ = 5 * 2x $^{2}$ = 10x $^{2}$

The coefficient is 10.

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Binomial Theorem

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Vraag 37 Verslag

A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.

17.55i + 13.78j

17.55j - 13.78i

-17.55i + 13.78j

-17.55i - 13.78j

Statics

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Vraag 38 Verslag

If α and β are roots of $x^2 + mx - n = 0$, where m and n are constants, form the

equation

whose

roots

are

1α

and

1β

.

mnx

^{2}

- n

^{2}

x - m = 0

mx

^{2}

- nx + 1 = 0

nx

^{2}

- mx + 1 = 0

nx

^{2}

- mx - 1 = 0

Antwoorddetails

x $^{2}$ + mx - n = 0

a = 1, b = m, c = -n

α + β = $\frac{- b}{a}$ = $\frac{- m}{1}$ = -m

αβ = $\frac{c}{a}$ = $\frac{- n}{1}$ = -n

the roots are = $\frac{1}{α}$ and $\frac{1}{β}$

sum of the roots = $\frac{1}{α}$ + $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ $\frac{α + β}{α β}$

α + β = -m
αβ = -n

$\frac{α + β}{α β}$

product of the roots = $\frac{1}{α}$ * $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ = $\frac{1}{α β}$ → $\frac{1}{- n}$

x $^{2}$ - (sum of roots)x + (product of roots)
x $^{2}$ - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx $^{2}$ - mx - 1 = 0

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Vraag 39 Verslag

In how many ways can six persons be paired?

5

10

15

20

Antwoorddetails

6C $_{2} = \frac{6!}{[6 - 2]! [2!]}$

$\frac{6 * 5 * 4!}{4! * 2!}$

= $\frac{6 * 5}{2}$

= 15

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Vraag 40 Verslag

Given that P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.

(10 units, 053º)

(9 units, 049º)

(10 units, 037º)

(9 units, 027º)

Vectors

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Vraag 41 Verslag

A straight line makes intercepts of -3 and 2 on the x and y axes respectively. Find the equation of the line.

2x + 3y + 6 = 0

3x - 2y - 6 = 0

-3x 2y - 6 = 0

-2x + 3y - 6 = 0

Co-ordinate Geometry

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Vraag 42 Verslag

A binary operation ∆ is defined on the set of real numbers R, by $x∆y = \sqrt{x+y-\frac{xy}{4}}$, where x, yER. Find the value of $4∆3$

16

8

4

2

Antwoorddetails

x∆y = $\sqrt{x + y - \frac{x y}{4}}$

4∆3 = $\sqrt{4 + 3 - \frac{4 * 3}{4}}$

= $\sqrt{4 + 3 - 3}$

= $\sqrt{4}$

= 2.

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Vraag 43 Verslag

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

What is the mode of the distribution?

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

5

6

7

8

Statistics

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Vraag 44 Verslag

The probability that a student will graduate from college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

0.06

0.22

0.78

0.80

Probability

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Vraag 45 Verslag

A particle initially at rest moves in a straight line with an acceleration of $ (10t - 4t^2)\text{ m/s}^2 $. Find the:

a. velocity of the particle after t seconds;

ii. average acceleration of the particle during the 4th second.

b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take $ g = 10\text{ m/s}^2 $]

ii. with an acceleration of $ 3\text{ m/s}^2 $. [Take $ g = 10\text{ m/s}^2 $]

Antwoord

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

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Antwoorddetails

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

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Vraag 46 Verslag

The vectors 6i + 8j and 8i - 6j are parallel to ?OP and ?OQ respectively. If the magnitude of ?OP and ?OQ are 80 units and 120 units respectively, express: ?OP and ?OQ in terms of i and j;

ii. |?PQ|, in the form c?k, where c and k are constants.

Antwoord

Vectors

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Vraag 47 Verslag

Solve $2^{(2y+1)} - 5(2^y) + 2 = 0$

Antwoord

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or $\frac{1}{2}$

p = 2 $^{y}$
when p = 2
2 $^{y}$ = 2
y = 1

when p = $\frac{1}{2}$

2 $^{y}$ = $\frac{1}{2}$

2 $^{y}$ = 2 $^{- 1}$

y = -1

y = -1 or 1

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Antwoorddetails

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or $\frac{1}{2}$

p = 2 $^{y}$
when p = 2
2 $^{y}$ = 2
y = 1

when p = $\frac{1}{2}$

2 $^{y}$ = $\frac{1}{2}$

2 $^{y}$ = 2 $^{- 1}$

y = -1

y = -1 or 1

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Vraag 48 Verslag

Two functions f and g are defined on the set of real numbers, R, by

$f:x\to x^2+2$ and $g:x\to \frac{1}{x+2}$.Find the domain of $(g\circ f)^{-1}$

Antwoord

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

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Antwoorddetails

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

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Vraag 49 Verslag

A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.

Two fruits are drawn one after the other without replacement. Calculate the probability that:

i. the first is an orange and the second is an avocado pear.

ii.both are of same fruit;

iii. at least one is an apple

Antwoord

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

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Antwoorddetails

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

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Vraag 50 Verslag

A body of mass of 18kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35º. If the system is in equilibrium, calculate the:

i. value of F

ii. tension in the string

Antwoord

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

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Antwoorddetails

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

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Statics

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Vraag 51 Verslag

A solid rectangular block has a base that measures 3x cm by 2x cm. The height of the block is ycm and its volume is 72cm$^3$.

i. Express y in terms of x.

ii. An expression for the total surface area of the block in terms of x only;

iii. the value of x for which the total surface area has a stationary value.

Antwoord

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

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Antwoorddetails

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

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Vraag 52 Verslag

Evaluate: $ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} $.

Antwoord

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

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Antwoorddetails

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

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Differentiation

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Vraag 53 Verslag

Given that p = (8N,030º) and q = (9N, 150º), find, in component from, the unit vector along(p - q).

Antwoord

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

Lees lesnotitie op Vectors (WAEC)

Antwoorddetails

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

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Vraag 54 Verslag

Given that $nC_4$, $nC_5$ and $nC_6$ are the terms of a linear sequence (A.P), find the :

i. value of n

ii. common differences of the sequence.

Antwoord

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

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Antwoorddetails

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

Lees lesnotitie op Sequences And Series (WAEC)

Lees lesnotitie op Permutation And Combinations (WAEC)

Sequences And Series

Permutation And Combinations

Bekijk antwoord

Vraag 55 Verslag

The table shows the scores obtained by a group of artistes in Vocal (X) and the instrument (Y) musical competition.

Vocal (X)	63	69	72	59	82	91	95	68
Instrument (Y)	58	61	67	51	53	79	92	57

Calculate the spearman's rank correlation coefficient between the scores.

Antwoord

p = $1 - \frac{6 Σ (d_{i})^{2}}{n (n^{2} - 1}$

p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation

X	R $_{x}$	Y	R $_{y}$	D $_{I}$ (R $_{x}$ - R $_{y}$ )	D $_{i}$ $^{2}$
63	2	58	4	-2	4
69	4	61	5	-1	1
72	5	67	6	-1	1
59	1	51	1	0	0
82	6	53	2	4	16
91	7	79	7	0	0
95	8	92	8	0	0
68	3	57	3	0	0

P = $1 - \frac{6 (4 + 1 + 1 + 0 + 16 + 0 + 0 + 0)}{8 (8^{2} - 1)}$

p = $1 - \frac{6 (22)}{8 (63)}$

p = $1 - \frac{132}{504}$

p = $1 - \frac{11}{42}$

p = $\frac{11}{42}$

Lees lesnotitie op Statistics (WAEC)

Antwoorddetails

p = $1 - \frac{6 Σ (d_{i})^{2}}{n (n^{2} - 1}$

p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation

X	R $_{x}$	Y	R $_{y}$	D $_{I}$ (R $_{x}$ - R $_{y}$ )	D $_{i}$ $^{2}$
63	2	58	4	-2	4
69	4	61	5	-1	1
72	5	67	6	-1	1
59	1	51	1	0	0
82	6	53	2	4	16
91	7	79	7	0	0
95	8	92	8	0	0
68	3	57	3	0	0

P = $1 - \frac{6 (4 + 1 + 1 + 0 + 16 + 0 + 0 + 0)}{8 (8^{2} - 1)}$

p = $1 - \frac{6 (22)}{8 (63)}$

p = $1 - \frac{132}{504}$

p = $1 - \frac{11}{42}$

p = $\frac{11}{42}$

Lees lesnotitie op Statistics (WAEC)

Statistics

Bekijk antwoord

Vraag 56 Verslag

The probability that Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:

(a) only be late for 3 days.

(b) not be late in the week:

(c) be late throughout the six days.

Antwoord

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

Lees lesnotitie op Probability (WAEC)

Antwoorddetails

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

Lees lesnotitie op Probability (WAEC)

Probability

Bekijk antwoord

Vraag 57 Verslag

Solve 3cos2x - sinx = 0 for 0º≤x≤360º

Antwoord

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

Lees lesnotitie op Trigonometry (WAEC)

Lees lesnotitie op Co-ordinate Geometry (WAEC)

Antwoorddetails

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

Lees lesnotitie op Trigonometry (WAEC)

Lees lesnotitie op Co-ordinate Geometry (WAEC)

Trigonometry

Co-ordinate Geometry

Bekijk antwoord

Vraag 58 Verslag

A binary operation * is defined on the set $T = \{-2,-1,1,2\}$ by $p*q = p^2 + 2pq - q^2$, where $p,q \in T$.
Copy and complete the table.

*	-2	-1	1	2
-2		7		-8
-1		2	-2
1	-7			1
2		-1

Antwoord

p*q = p $^{2}$ + 2pq - q $^{2}$

when p = -2 and q = -2

p*q = -2 $^{2}$ + 2(-2)(-2) - (-2) $^{2}$
p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1
p*q = -2 $^{2}$ + 2(-2)(-1) - (-1) $^{2}$
p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1
p*q = -2 $^{2}$ + 2(-2)(1) - (1) $^{2}$
p*q = 4 - 4 -1 = -1

when p = -1 and q = -2
p*q = -1 $^{2}$ + 2(-1)(-2) - (-2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1
p*q = -1 $^{2}$ + 2(-1)(-1) - (-1) $^{2}$
p*q = 1 + 2 -1 = 2

when p = -1 and q = 2
p*q = -1 $^{2}$ + 2(-1)(2) - (2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2
p*q = 1 $^{2}$ + 2(1)(-2) - (-2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1
p*q = 1 $^{2}$ + 2(1)(-1) - (-1) $^{2}$
p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1
p*q = 1 $^{2}$ + 2(1)(1) - (1) $^{2}$
p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2
p*q = 1 $^{2}$ + 2(1)(2) - (2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2
p*q = 2 $^{2}$ + 2(2)(-2) - (-2) $^{2}$
p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1
p*q = 2 $^{2}$ + 2(2)(-1) - (-1) $^{2}$
p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1
p*q = 2 $^{2}$ + 2(2)(1) - (1) $^{2}$
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 2 $^{2}$ + 2(2)(2) - (2) $^{2}$
p*q = 4 + 8 - 4 = 8

*	-2	-1	1	2
-2	8	7	-1	-8
-1	1	2	-2	-7
1	-7	-2	2	1
2	-8	-1	7	8