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\(\int (2x + 3)^{\frac{1}{2}} \delta x\) let u = 2x + 3, \(\frac{\delta y}{\delta x} = 2\) \(\delta x = \frac{\delta u}{2}\) Now \(\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}\) \( = \frac{1}{2} \int u^{\frac{1}{2}} \delta u\) \( = \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k\) \( = \frac{1}{3} u^{\frac{3}{2}} + k\) \( = \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k\)

Mean x = \(\frac{\sum x}{n}\) = [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1)] \(\div\) 5 = [11 - 1 + 3t - 3t] \(\div\) 5 = 10 \(\div\) 5 = 2

The number with the highest frequency = 4

First arrange the numbers in order of magnitude; 1,2,3,3,4,5,5,5,5,6,7,8,9,9,10 Hence the median = 5

Mean x = \(\frac{\sum x}{n}\) \( = \frac{5 + 4 + 3 + 2 + 1}{5}\) \( = \frac{15}{5}\) = 3 \(\begin{array}{c|c} x & d = x - 3 & d^2 \\ \hline 5 & 2 & 4 \\ 4 & 1 & 1 \\ 3 & 0 & 0 \\ 2 & -1 & 1 \\ 1 & -2 & 4 \\ \hline & & \sum d^2 + 10 \end{array}\) Hence, standard deviation; \( = \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}}\) \( = \sqrt{2}\)

A team of 2 girls can be selected from 7 girls in \(^7C_3\) \( = \frac{7!}{(7 - 3)! 3!}\) \( = \frac{7!}{4! 3!} ways\)

Let E demote the event of obtaining at least a 4 Then n(E) = 16 + 10 + 14 = 40 Hence, prob (E) = \(\frac{n(E)}{n(S)}\) \( = \frac{40}{100}\) \( = \frac{2}{5}\)

Sample space S = {10, 11, 12, ... 30} Let E denote the event of choosing a number divisible by 3 Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7 Prob (E) = \(\frac{n(E)}{n(S)}\) Prob (E) = \(\frac{7}{21}\) Prob (E) = \(\frac{1}{3}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

In the diagram above, \(\alpha\) = 54o(alternate angles; KL||MN) o 35o + o = 180o(sum of angles of \(\bigtriangleup\)) o = 180o o - 143o = 37o

In the figure above, qo = 30o (vertically opposite angles) (P + 2q)o + 30o = 180o(angles on a straight line) p + 2 x 30o + 30o = 180o p + 60o + 30o = 180o p + 90o = 180o p = 180o - 90o = 90o

In the figure above, \(\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}\) (Sine rule) x = \(\frac{10 \sin 60^o}{\sin 30^o}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\) = 10 x \(\frac{\sqrt{3}}{2} \times \frac{2}{1}\) = 10\(\sqrt{3}\)cm

(x1, y1) = (0,5) (x2, y2) = (5, 0) Using \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0}\) \(\frac{y - 5}{-5} = \frac{x}{5}\) 5(y - 5) = -5x y - 5 = -x x + y = 5 y = -x + 5

Angle of sector subtended by yam = 360o - (70 + 80 + 50)o = 360o - 200o = 160o But \(\frac{80^o}{360^o}\) x T = 8000 T = \(\frac{8000 \times 360^o}{80^o}\) = N36,000 Hence the amount spent on yam = \(\frac{160^o}{260} \times N36,000\) = N16,000

10\(\frac{2}{5} - 6 \frac{2}{3} + 3\) \(\frac{52}{5} - \frac{20}{3} + \frac{3}{1}\) = \(\frac{156 - 100 + 45}{15}\) \(\frac{156 + 45 - 100}{15}\) = \(\frac{201 - 100}{15}\) = \(\frac{101}{15}\) = 6\(\frac{11}{15}\)