Further Mathematics Questions & Answers - Free Online Practice

216.

The equation of the line of best fit for variables x and y is $y = 19.33 + 0.42x$ , where x is the independent variable. Estimate the value of y when x = 15.

18.91

19.75

25.63

38.23

View answer & explanation

Correct answer is C

$y = 19.33 + 0.42x$

$\text{The value of y when x = 15} = 19.33 + (0.42 \times 15)$

= $19.33 + 6.30$

= 25.63

217.

In a firing contest, the probabilities that Kojo and Kwame hit the target are $\frac{2}{5}$ and $\frac{1}{3}$ respectively. What is the probability that none of them hit the target?

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{4}{5}$

View answer & explanation

Correct answer is B

No explanation has been provided for this answer.

218.

Age(in years)	1 - 5	6 - 10	11 - 15
Frequency	3	5	2

Calculate the standard deviation of the distribution.

1.10

2.36

3.50

7.50

View answer & explanation

Correct answer is C

Age (years)	Freq (f)	Mid-value (x)	fx	$d = (x - \bar{x})$	$d^{2}$	$fd^{2}$
1 - 5	3	3	9	- 4.5	20.25	60.75
6 - 10	5	8	40	0.5	0.25	1.25
11 - 15	2	13	26	5.5	30.25	60.5
$\sum =$	10		75			122.5

$Mean (\bar {x}) = \frac{\sum fx}{\sum f} = \frac{75}{10} = 7.5$

$SD = \sqrt{\frac{\sum fd^{2}}{\sum f}}$

= $\sqrt{\frac{122.5}{10}}$

= $\sqrt{12.25} = 3.50$

219.

A particle of mass 2.5 kg is moving at a speed of 12 m/s. If a force of magnitude 10 N acts against it, find how long it takes to come to rest.

1.5 s

3.0 s

4.0 s

6.0 s

View answer & explanation

Correct answer is B

$F = ma$

$10 = 2.5a \implies a = 4 ms^{-2}$

Since it is a retarding movement, then $a = -4 ms^{-2}$ .

$v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}$

$0 = 12 + (-4t) \implies 0 = 12 - 4t$

$4t = 12 \implies t = 3 s$

220.

The area of a sector of a circle is 3 $cm^{2}$ . If the sector subtends an angle of 1.5 radians at the centre, calculate the radius of the circle

1 cm

$\sqrt{2}$ cm

2 cm

4 cm

View answer & explanation

Correct answer is C

The area of sector in radians = $\frac{r^{2} \theta}{2}$

$3 cm^{2} = \frac{1.5 \times r^{2}}{2}$

$r^{2} = \frac{3 \times 2}{1.5} = 4$

$r = 2 cm$

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