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Score 0 1 2 3 4 5 Total Freq 2 3 4 2 7 2 20 \(fx\) 0 3 8 6 28 10 55 \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{55}{20}\) = \(2.75\)

There are 20 students. Hence, the median is found by the average of the 10th and 11th score. = \(\frac{3 + 3}{2}\) = 3

cos 60^\circ = 1/2, cos(180^\circ - 60^\circ) = -1/2 cos120^\circ = -1/2

From the diagram: PQ = 3 ; PR = 3+2 → 5 PQ: PR = \(\frac{3}{5}\)

cos 64^\circ = x/9 x = 9 cos 64^\circ =3.94 ≅ 4.0m

Let \(\theta\) be the angular difference between P (50^\circN. 50^\circW), and Q (50^\circN, 150^\circW), \(\theta\) = 150^\circ - 50^\circ =100^\circ

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

PQ\(^2\) = 24\(^2\) + 7\(^2\) PQ\(^2\) = 625 PQ = 25

Using Pythagoras' theorem: Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\) Hyp = \(\sqrt{625}\) → 25 \(\tan P = \frac{7}{24}\) tan P = 0.2917 P = tan\(^{-1}\) (0.2917) = 16.26^\circ Bearing of Q from P = 180^\circ - 16.26^\circ = 163.74^\circ \(\approxeq\) 164^\circ

sinθ = 3/5 tanθ = 3/4

(271/3)2 = (33 x 1/3)2 = 32 = 9

The linear equation will be the difference between the equation. y=x2 + 4x - 6 and x2 + 4x - 7 = 0 y=x2 + 4x - 6, ( 0 = x2 + 4x - 7) y = -(-1) y = 1

vertically opposite angles are equal, hence 3x + 6x + 3x = 180^\circ 12x = 180^\circ x = 15^\circ

x + 2 and x - 7 are factors (x+2)(x+7) = 0 x(x-7) + 2(x-7) = 0 x2 - 7x + 2x - 14 = 0 x2 - 5x - 14 = 0

Error = 1.25 - 1.15 = 0.10 0.10/1.25* 100 = 8.0%