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Using \(V = \pi r^2 h\) 6160 = 22/7 x 28 x 28 x h \(h = \frac{6160}{22 \times 4 \times 28} \) \(h = 2.5m \)

Mid point of S(-5, 4) and T(-3, -2) is \([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + (-2))]\) \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(-8), \frac{1}{2}(2)]\) \([-4, 1]\)

\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{1}{2} = \frac{2 - 4}{1 - x}\) 1 - x = 2(2 - 4) 1 - x = 4 - 8 1 - x = -4 -x = -4 - 1 x = 5

y - 4x + 3 = 0 When y = 0, 0 - 4x + 3 = 0 Then -4x = -3 x = 3/4 So the line cuts the x-axis at point (3/4, 0). When x = 0, y - 4(0) + 3 = 0 Then y + 3 = 0 y = -3 So the line cuts the y-axis at the point (0, -3) Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\) \([\frac{3}{8}, \frac{-3}{2}]\)

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

If y = 4x3 - 2x2 + x, then; \(\frac{\delta y}{\delta x}\) = 3(4x2) - 2(2x) + 1 = 12x2 - 4x + 1

y = cos 3x Let u = 3x so that y = cos u Now, \(\frac{\delta y}{\delta x} = 3\), \(\frac{\delta y}{\delta x} = -sin u\) By the chain rule, \(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\) \(\frac{\delta y}{\delta x} = (-\sin u) (3)\) \(\frac{\delta y}{\delta x} = -3 \sin u\) \(\frac{\delta y}{\delta x} = -3 \sin 3x\)

y = x2 - 2x - 3, Then \(\frac{\delta y}{\delta x} = 2x - 2\) But at minimum point,\(\frac{\delta y}{\delta x} = 0\), Which means 2x - 2 = 0 2x = 2 x = 1. Hence the minimum value of y = x2 - 2x - 3 is; ymin = (1)2 - 2(1) - 3 ymin = 1 - 2 - 3 ymin = -4

\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\) \(- \frac{1}{2} \cos 2x + k\)

\(\int (2x + 3)^{\frac{1}{2}} \delta x\) 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 \(^7C_3\) \( = \frac{7!}{(7 - 3)! 3!}\) \( = \frac{7!}{4! 3!} ways\)