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Given \(\sin x = 0.6 = \frac{3}{5}\) \(\therefore Opp = 3; Hyp = 5\) Using Pythagoras theorem, \(5^{2} = 3^{2} + Adj^{2}\) \(Adj = \sqrt{5^{2} - 3^{2}}\) \(Adj = \sqrt{25 - 9} = \sqrt{16} = 4\) \(\therefore \cos x = \frac{4}{5} = 0.8\) \(2\cos x + 3\sin x = 2(\frac{4}{5}) + 3(\frac{3}{5})\) = \(\frac{8}{5} + \frac{9}{5}\) = \(\frac{17}{5}\). (b) Join O to X and O to Y. Considering either \(\Delta OXC\) or \(\Delta OYC\), \(\sin 65 = \frac{r}{30}\) \(r = 30 \sin 65\) \(r = 30 \times 0.9063\) \(r = 27.189 cm \approxeq 27.2 cm\) (to one decimal place) (b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ. = \(\frac{1}{2} bh - \frac{1}{2} \pi r^{2}\) = \((\frac{1}{2} \times 2c \times 30) - (\frac{1}{2} \times \frac{22}{7} \times (27.189)^{2})\) From \(\Delta COA, \tan 65 = \frac{c}{30}\) \(\implies c = 30 \tan 65 \approxeq 64.34 cm\) \(\therefore\) Area of shaded portion = \(30(64.34) - \frac{8131.659}{7}\) = \(1930.056 - 1161.666\) = \(768.39 cm^{2}\)

(c) |XP| = 3.8 cm.

(a) Total surface area of a sphere = \(4\pi r^{2}\) Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R. \(\frac{S_{1}}{S_{2}} = \frac{9}{49}\) \(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\) \(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\) \(\frac{49 \times 144}{9} = R^{2}\) \(R = \sqrt{49 \times 16} = 28 cm\) Volume of bigger sphere = \(\frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 28 \times 28 \times 28\) = \(\frac{275,968}{3}\) = \(91989.33 cm^{3}\) \(\approxeq 91989 cm^{3}\) (nearest whole number). (b) \(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \cos Y\) = \(5^{2} + 3^{2} - 2(5)(3) \cos 135\) = \(25 + 9 - 30(-0.7071)\) = \(34 + 21.213\) \(|ZX|^{2} = 55.213\) \(|ZX| = 7.43 km \approxeq 7 km\) (ii) Using sine rule, \(\frac{\sin \theta}{5} = \frac{\sin 135}{7.43}\) \(\sin \theta = \frac{5 \times \sin 135}{7.43}\) = \(\frac{3.5355}{7.43}\) \(\sin \theta = 0.4758\) \(\theta = \sin^{-1} (0.4758)\) = \(28.41^\circ \approxeq 28^\circ\) (to the nearest degree). Bearing of Z from X = 270^\circ + 28.41^\circ = 298.41^\circ \(\approxeq\) 298^\circ.

Score 1 2 3 4 5 6 Frequency 2 5 x 11 9 10 P(3) = 0.26 \(\frac{x}{37 + x} = 0.26\) \(x = 0.26(37 + x) \implies x = 9.62 + 0.26x\) \(x - 0.26x = 9.62 \implies 0.74x = 9.62\) \(x = \frac{9.62}{0.74} = 13\) Total toss frequency = 37 + x = 37 + 13 = 50. (a) Median = \(\frac{n + 1}{2}\) = \(\frac{50 + 1}{2}\) = 25.5 From the table, 25.5 occurs at 4 which is the median. (b) Standard deviation = \(\sqrt{\frac{\sum fd^{2}}{\sum f}}\) \(x\) \(f\) \(\fx\) \(d = |x - \bar{x}|\) \(d^{2}\) \(fd^{2}\) 1 2 2 -3 9 18 2 5 10 -2 4 20 3 13 39 -1 1 13 4 11 44 0 0 0 5 9 45 1 1 9 6 10 60 2 4 40 \(\sum\) 50 200 100 \(Mean(\bar{x}) = \frac{\sum fx}{\sum f}\) = \(\frac{200}{50} = 4 S.D = \(\sqrt{\frac{100}{50}}\) = \(\sqrt{2}\) = 1.4142

(a) \(\theta = \frac{60^\circ}{2} = 30^\circ\) Area of trapezium = \(\frac{a + b}{2} h\) \(60 = \frac{15 + 25}{2} h \implies 60 = 20h\) \(h = \frac{60}{20} = 3 cm\) (ii) Considering \(\Delta PSR\), \(\sin \theta = \frac{Opp}{Hyp}\) \(\sin 30 = \frac{3}{|PS|}\) \(|PS| = \frac{3}{\sin 30}\) = \(\frac{3}{0.5} = 6 cm\) (b) Let the total amount to be shared = k. Ade's share = \(\frac{3}{5} k \) Remainder = \(k - \frac{3}{5}k \) = \(\frac{2}{5}k\) Nelly's share : \(\frac{1}{3} \times \frac{2}{5}k\) = \(\frac{2}{15}k\) Austin's share : \(\frac{2}{5}k - \frac{2}{15}k\) = \(\frac{4}{15}k\) \(\frac{4}{15}k - \frac{2}{15}k = \frac{2}{15}k\) \(\frac{2}{15}k = N3,000\) \(\implies k = \frac{3,000 \times 15}{2}\) = \(N22,500\) Ade's share = \(\frac{3}{5}k\) = \(\frac{3}{5} \times N22,500 = N13,500\).

(a) \(P \propto Q\) and \(P \propto \frac{1}{R^{2}}\). \(P \propto \frac{Q}{R^{2}} \implies P = \frac{kQ}{R^{2}}\) When P = 1, Q = 8, R = 2 \(1 = \frac{8k}{2^{2}} \implies 8k = 4\) \(k = \frac{4}{8} = 0.5\) \(P = \frac{Q}{2R^{2}}\) When P = 3, R = 5, Q = ? \(3 = \frac{Q}{2(5^{2})}\) \(3 = \frac{Q}{50} \implies Q = 150\). (b) Angular difference : 60^\circ - 20^\circ = 40^\circ \(\frac{\theta}{360} \times 2\pi r \) \(r = R\cos \theta\) \(\frac{40}{360} \times 2 \times 3.142 \times 6400 \cos 20 = \frac{1511687.278}{360}\) = \(4199.13 km\) \(Speed = \frac{Distance}{Time}\) = \(\frac{4199.13}{6}\) = \(699.85 km/hr\) \(\approxeq 700 km/hr\) (to 3 significant figures) (ii) \(D = \frac{\theta}{360} \times 2\pi r\) \(420 = \frac{\theta}{360} \times 2 \times 3.142 \times 6400\) \(\theta = \frac{420 \times 360}{2 \times 3.142 \times 6400}\) \(\theta = \frac{151200}{40217.6} = 3.76^\circ\) Latitude of C : 20^\circ + 3.76^\circ = 23.76^\circ Note that due North implies same longitude but different latitude.

Median = \(\frac{\text{N}}{2}\) th item = \(\frac{150}{2}\) th ≈ 36.5 from the graph.

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

Inception; the commencement or starting point of something

Detached: separate or disconnected.

Dwelt; to live in or at a specified place.

Ignominious; deserving or causing public disgrace or shame.

compliments; a polite expression of praise or admiration.

Ensnared; to trap or gain power over (someone) by dishonest or underhand means.

corrugated; (of a material or surface) shaped into a series of parallel ridges and grooves so as to give added rigidity and strength. Folded