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circumference of the bicycle wheel \(= 2\pi r = 2 \times \frac{22}{7} \times 42 = 264cm = 2.64m\\\) Number of revolution \(=\frac{66}{2.64} = 25\)

Length of arc = \(\frac{\theta}{360} \times 2\pi r\) \(50 = \frac{75}{360} \times 2 \times \frac{22}{7} \times r\) \(r = \frac{50 \times 360 \times 7}{75 \times 2 \times 22}\) \(r = \frac{420}{11}\) = 38.18 cm \(\approxeq\) 38.2 cm (3 sig. figs)

From the figure above, and by measurement < RMQ = 60^\circ.

In triangle ABC, \(\frac{BC}{AC} = \cos \theta\) \(\implies \cos \theta = \frac{5}{10} = 0.5\) \(\theta = \cos^{-1} (0.5) = 60^\circ\)

(a) see figure a above \(\frac{h}{100} = \sin 24\) \(h = 100 \sin 24\) = \(100 \times 0.4067\) = \(40.67m\) (b) see figure b above \(\sin \alpha = \frac{4}{5} = 0.8\) \(\alpha = \sin^{-1} (0.8) = 53.13^\circ\) \(< BAC = 2 \theta = 2(53.13^\circ)\) = \(106.26^\circ\) \(\approxeq 106^\circ\) (to the nearest degree). (c) See Figure C above In \(\Delta PSQ\), \(\frac{h}{PQ} = \tan 71.6\) \(PQ = \frac{h}{\tan 71.6}\) In \(\Delta QSR\), \(\frac{h}{QR} = \tan 45\) \(QR = \frac{h}{\tan 45}\) \(PR = PQ + QR\) \(\frac{h}{3} + h = 70 \implies \frac{4}{3}h = 70\) \(h = \frac{70 \times 3}{4} = 52.5m\)

\(x^2 = 8^2 + 8^2\) \(x^2 = 64 + 64 = 128\) \(x = \sqrt{128}\) = \(8\sqrt{2}\) km

The bearing of Kofi from the building = 180^\circ + 45^\circ = 225^\circ

S50^\circW = 180^\circ + 50^\circ = 230^\circ

The bearing of the ship from its original position is S90ºE

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\)16 - x\(^2\)dx = [16 - \(\frac{x^3}{3} + c]^{4}_{0}\) = \(\frac{128}{3}\) square units

Students that scored 4 and above = 5 + 3 + 3 + 0 + 2 + 2 + 2 = 17

2 + 3 + 4 + 5 + 6 + 3 +3 + 0 + 2 + 2 + 2 = 32

If the total number of students represents 100%, then 70% corresponds to the point on the graph where the cumulative frequency reaches 70 From the graph, find where the cumulative frequency line intersects the 70% mark. This typically corresponds to a specific age on the X-axis

(a)(i) The number of pupils who scored between 41 and 50 marks = \(\frac{90}{360} \times 200 = 50\) (ii) The angles for pupils above 60 marks = 54^\circ + 36^\circ + 18^\circ = 108^\circ \(\frac{108}{360} \times 200 = 60\) (b) The fraction of the pupils who scored at least 50 marks : \(\frac{54 + 90}{360} = \frac{2}{5}\) (c) The modal class = 51 - 60; It has the largest angle in the pie- chart.

x 2 3 6 2 5 0 4 2 24 x - \(\bar{x}\) -1 0 3 -1 2 -3 1 -1 \((x - \bar{x})^2\) 1 0 9 1 4 9 1 1 26 Mean = \(\frac{24}{8} = 3\) Standard deviation = \(\sqrt{\frac{26}{8}}\) = \(\sqrt{3.25}\) = 1.802 \(\approx\) 1.8