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2y + Reflex angle = 360 Reflex angle = 360 - 2y 360 - 2y = 2(106)(angles at a centre = 2ce < of circum.) 360 - 212 = 2y 148 = 2y y \(\frac{148}{2}\) y = 74o

Area of shaded portion = Area of semicircle Area of \(\bigtriangleup\) RSO Area of semicircle = \(\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 2}\) = 25.14cm2; Area of \(\bigtriangleup\)RSO =\(\sqrt{s(s - 1)(s - b)(s - c)}\); where s = \(\frac{a + b + c}{2}\) s = \(\frac{4 + 4 + 4}{2}\) = 6cm = \(\sqrt{6(6 - 4)(6 - 4) (6 - 4)}\) = \(\sqrt{6(2) (2) (2)}\) = \(\sqrt{18}\) = 6.93cm2 Area of shaded region = 25.14 - 6.93 = 18.21cm2

(a) \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\) = \(\frac{3.69}{1.64} \times 10^{5 - (-3)}\) = \(2.25 \times 10^{8}\) (b) Interest from bank A : \(\frac{y}{100} \times N20,000 = N200y\) Interest on bank B :\(\frac{1.5y}{100} \times N25,000 = N375y\) \(\therefore N(200y + 375y) = N4,600\) \(575y = 4,600 \implies y = 8\)

From \(\Delta\) BCD, \(\sin \theta = \frac{50}{56}\) \(\theta = \sin^{-1} (0.8929)\) \( < BCD = 63.23^\circ\) \(\therefore < ACB = 63.23^\circ \times 2 = 126.46^\circ\) \(< ACB (reflex) = 360^\circ - 126.46^\circ\) = \(233.54^\circ\) Perimeter of the arc = \(\frac{233.54}{360} \times 2 \times \frac{22}{7} \times 56\) = \(228.35 m\) Perimeter of the railway tunnel = Perimeter of the arc + |AB| = 228.35 m + 100 m = 328.35 m.

The diagram above is the solution

(a) \(\frac{x^{2} - y^{2}}{3x + 3y}\) \(\frac{(x + y)(x - y)}{3(x + y)}\) (Using difference of two squares) = \(\frac{x - y}{3}\) (b)(i) \(\sin 37 = \frac{/PS/}{15}\) \(/PS/ = 15 \times 0.6018\) = \(9.03 cm\) (ii) \(\cos 37 = \frac{/SK/}{15}\) \(/SK/ = 15 \times 0.7986\) = \(11.98 cm \) \(\approxeq 12.0 cm\) (iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK. /SR/ = 2(/SK/) = 2(11.98) = 23.96 cm Area of rectangle PQRS = \(23.96 \times 9.03 \) = \(216.3588 cm^{2}\) Area of triangle PKS = \(\frac{1}{2} \times 11.98 \times 9.03\) = \(54.088 cm^{2}\) Area of shaded portion : \((216.3588 - 54.088)cm^{2}\) \(162.2708 cm^{2}\) \(\approxeq 162 cm^{2}\)

\(x\) mid-value \(f\) \(fx\) \((x - \bar{x})\) \((x - \bar{x})^{2}\) \(f(x - \bar{x})^{2}\) 0 - 4 2 1 2 -12 144 144 5 - 9 7 5 35 -7 49 245 10 - 14 12 10 120 -2 4 40 15 - 19 17 9 153 3 9 81 20 - 24 22 5 110 8 64 320 \(\sum \) 30 420 830 (a) Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{420}{30} = 14\) (b) Standard deviation = \(\sqrt{\frac{\sum f(x - \bar{x})^{2}}{\sum f}\) = \(\sqrt{\frac{830}{30}}\) = \(\sqrt{27.66}\) = 5.26.