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Mode = a + \(\frac{ (b - a)(fm - Fb)}{2Fm - Fa - Fb}\) = 3.0 + \(\frac{(3.45 - 2.95)(15 - 4)}{2(15) - 4 - 10}\) = 3 + \(\frac{(0.5)(11)}{30 - 14}\) = 3 + \(\frac{5.5}{16}\) = 3 + 0.34375 = 3.34375 = 3.3cm

Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\)) = 2.95 + \(\frac{0.5}{15}\)(2.-7) = 2.95 + \(\frac{0.5}{15}\) x 13 = 2.95 + 0. 43 = 3.38 = 3.4

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

\(\begin{array}{c|c} & W & W & W & W & B & B \\ \hline W & WW & WW & WW & WW & WB & WB \\ W & WW & WW & WW & WW & WB & WB\\W & WW & WW & WW & WW & WB & WB\\ W & WW & WW & WW & WW & WB & WB\\ B & BW & BW & BW & BW & BB & BB \\ B & BW & BW & BW & BW & BB & BB\end{array}\) P(WW) = \(\frac{16}{36}\) = \(\frac{4}{9}\)

-1 \(\geq\) x \(\geq\) 1, 1 < x \(\geq\) 2 Combining solutions = x \(\leq\) 1; 1 \(\geq\) x \(\geq\) 2

Within the triangle, 180^\circ - (81^\circ + 53^\circ) = 46^\circ In a cyclic quadrilateral, the sum of two opposite angles = 180^\circ \(\therefore\) x = 180^\circ - 46^\circ = 134^\circ

XTY = 50o = TXY = \(\frac{130^o}{2}\) = 65o

Area(S) = 2\(\pi r^2 + 2 \pi rh = 2\pi r(h + r)\) D = 14cm, 2r = D r = \(\frac{D}{2}\) = 7cm S = 2 x \(\frac{22}{7} \times [7 + 12] = 44 \times 19\) = 836cm2

\(\frac{1}{2}\)(PQ x 10 sin 30) = 35 PQ = \(\frac{35 \times 2}{10 \sin 30}\) = 14cm

Given:1.20, 1.00, 0.90, 1.40, 0.80, 1.20, 1.10 Sum of the numbers = 1.20 + 1.00 + 0.90 + 1.40 + 0.80 + 1.20 + 1.10 = 7.60 Number of terms = 7 Mean, m = 7.60 \div 7 = 1.0857 (to 4 d.p.) Highest value = 1.40 Lowest value = 0.80 Range, r = 1.40 − 0.80 = 0.60 m + r = 1.0857 + 0.60 = 1.6857 closest to 1.65.

(a) \(px - 2qx - 4qy + 2py\) = \(x(p - 2q) + 2y(p - 2q)\) = \((x + 2y)(p - 2q)\) (b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\) \(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\) \((P \cup Q) = {1, 2, 3, 4, 6, 9, 10}\) \((P \cup Q)' = {5, 7, 8}\) \(P' \cap Q' = {3, 5, 7, 8, 9} \cap {1, 4, 5, 7, 8, 10}\) = \({5, 7, 8}\) Hence, \((P \cup Q)' = P' \cap Q'\).

(a) \(y = k + \frac{c}{x^{2}}\) where c and k are constants. (b) When x = 1, y = 11 \(11 = k + \frac{c}{1^{2}} \implies 11 = k + c ... (1)\) When x = 2, y = 5 \(5 = k + \frac{c}{2^{2}} \implies 5 = k + \frac{c}{4}\) \(\equiv 20 = 4k + c ... (2)\) (2) - (1) : \(4k - k = 20 - 11 \implies 3k = 9\) \(k = 3\) Put k = 3 in (1), we have \(11 = 3 + c \implies c = 11 - 3 = 8\) \(\therefore y = 3 + \frac{8}{x^{2}}\) When x = 4, \(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\) = \(3\frac{1}{2}\)

(a) \(/QY/ = 2cm; /RS/ = 3cm\) \(\therefore /YZ/ = /RS/ = /PQ/ = 3cm\) (opp sides of a parallelogram) But \(/PZ/ = /PQ/ + /QY/ + /YZ/\) \(\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\) (b) Difference in longitude = 95^\circ - 25^\circ = 70^\circ Distance PQ = \(\frac{70}{360} \times 2\pi R \cos \theta\) = \(\frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 56\) = \(\frac{7}{36} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592\) = \(4374.19 km \approxeq 4374 km\) (to the nearest km)

(a) Total no of cards = 52; No of fives = 4; No of red nines = 2. Prob(of 5) = \(\frac{4}{52} = \frac{1}{13}\) Prob(red nine) = \(\frac{2}{52} = \frac{1}{26}\) \(\therefore\) Prob (of 5 or red nine) = \(\frac{1}{13} + \frac{1}{26} = \frac{3}{26}\) (b) \(\tan < RPQ = \frac{3}{5} = 0.6\) \(< RPQ = 30.96^\circ\) \(\therefore\) Bearing of R from P = 150^\circ - 30.96^\circ = 119.04^\circ \(\approxeq\) 119^\circ (to the nearest degree).

x 1 2 3 4 5 6 7 8 9 10 Total f 2 3 4 4 4 4 5 3 2 1 32 fx 2 6 12 16 20 24 35 24 18 10 167 (a)(i) Mean = \(\frac{\sum fx}{\sum f} = \frac{167}{32}\) = \(5.21875 \approxeq 5.22\) (ii) By inspection, the average of the numbers in the 16th and 17th position give the median = \(\frac{5 + 5}{2} = 5\) (iii) Mode = 7 marks having the largest frequency. (b) Students who scored at least 8 marks = Those who scored 8, 9, 10. Total number of students that scored at least 8 = 3 + 2 + 1 = 6. \(%age = \frac{6}{32} \times 100% = 18.75%\)