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

Given the class intervals: \(3 - 7\), \(8 - 12\), \(13 - 17\), and \(18 - 22\): The class interval is \(4\): \(7 - 3 = 4 \quad \text{(True)}\) Class boundaries are \(2.5, 7.5, 12.5, 18.5, 22.5\): - Correct boundaries: \(2.5, 7.5, 12.5, 17.5, 22.5\) False. Mid-marks are \(5, 10, 15, 20\): \(\text{Mid-mark} = \frac{\text{Lower} + \text{Upper}}{2}\) - \(3 - 7: 5\) - \(8 - 12: 10\) - \(13 - 17: 15\) - \(18 - 22: 20\) True.

T = \(\frac{ \text{KD}}{\text{A}}\) 6 = \(\frac{3K}{2}\) 3K = 12 K = 4 T= \(\frac{4D}{A}\) but T = 3, D = 6 and A =? T = \(\frac{4D}{A}\) A = \(\frac{4D}{T}\) A = \(\frac{4 \times 6}{3}\) = 8

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

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

y\(^2\)+ 14y + k let a = 1, b = 14, c = k (where a and b represent the coefficient of y and y\(^2\) respectively) 4ac = b\(^2\) 4 x 1 x c = 14 x 14 = 196 4c = 196 c = \(\frac{196}{4}\) = 49

(a) \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\) = \(\log_{10} (\sqrt{\frac{30}{6}} \times \sqrt{2})\) = \(\log_{10} (\sqrt{5} \times \sqrt{2})\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} (10^{\frac{1}{2}})\) = \(\frac{1}{2} \log_{10} 10\) = \(\frac{1}{2}\). (b) (ii) \(A \cap C = { } \) (iii) \(A \cap B' = {1, 4}\).

(a) \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\) = \((\frac{5}{8} \times \frac{5}{2}) - (\frac{3}{4} \div \frac{3}{5})\) = \((\frac{25}{16}) - (\frac{5}{4})\) = \(\frac{25 - 20}{16}\) = \(\frac{5}{16}\) (b) \(V = \frac{1}{3} \times A \times h = \frac{1}{3} \pi r^{2} h\) \(\therefore \frac{1}{3} \times 154 \times h = \frac{1}{3} \times \pi \times r^{2} \times h\) Comparing the two equations, \(154 = \frac{22}{7} \times r^{2}\) \(r^{2} = \frac{154 \times 7}{22}\) \(r^{2} = 49\) \(\therefore r = 7 cm\).

(a) \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\) \(\frac{0.18 \times 12.5}{0.05 \times 0.2} = \frac{18 \times 10^{-2} \times 125 \times 10^{-1}}{5 \times 10^{-2} \times 2 \times 10^{-1}}\) = \(9 \times 25 \times 10^{-3 - (-3)}\) = \(9 \times 25\) = \(225\) \(\therefore \sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}} = \sqrt{225}\) = \(15\). (b) \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\) \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\) \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) \(\frac{8 - 4(3\sqrt{2})}{5\sqrt{2}} = \frac{8 - 12\sqrt{2}}{5\sqrt{2}}\) Rationalising, we have = \(\frac{8 - 12\sqrt{2}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{8\sqrt{2} - 24}{10}\) = \(0.8\sqrt{2} - 2.4\) = \(-2.4 + 0.8\sqrt{2}\) (c) \(x \propto \frac{y^{3}}{z^{2}}\) \(\implies x = \frac{ky^{3}}{z^{2}}\) \(108 = \frac{k \times 3^{3}}{4^{2}}\) \(k = \frac{108 \times 16}{27}\) \(k = 64\) \(\therefore x = \frac{64y^{3}}{z^{2}}\) \(\therefore 4000 = \frac{64 \times 10^{3}}{z^{2}}\) \(4 = \frac{64}{z^{2}}\) \(z^{2} = \frac{64}{4} = 16\) \(z = \sqrt{16} = \pm 4\)

(a) In \(\Delta ACB, < CAB = 180^\circ - (90^\circ + 61^\circ)\) = \(180^\circ - 151^\circ\) = \(29^\circ\) In \(\Delta ADB, < DAB = (29 + x)^\circ\) \(\cos (29 + x) = \frac{3}{5} = 0.6\) \((29 + x)^\circ = \cos^{-1} (0.6)\) \((29 + x)^\circ = 53.13^\circ\) \(x = 53.13^\circ - 29^\circ\) = \(24.13^\circ\) \(\approxeq 24.1^\circ\) (1 decimal place). (b) (i) \(DB = 2DN = 2NB\) \(OA = OB = OC = OD = 4cm\) \(AB = BC = CD = DA = 2cm\) \(DB^{2} = BC^{2} + DC^{2}\) = \(2^{2} + 2^{2}\) = \(\sqrt{8}\) = \(2\sqrt{2} cm\) \(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\) In \(\Delta DON\), \(OD^{2} = ON^{2} + DN^{2}\) \(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\) \(h^{2} = 14 \implies h = \sqrt{14}\) (ii) \(V = \frac{1}{3} \times \base area \times \perp distance\) \(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\) = \(\frac{4\sqrt{14}}{3}\) = \(4.99 cm^{3}\)

To divide \(6x^2 - 13x + 5\) by \(2x - 1\): \(\begin{array}{r|l} 3x - 5 & \\ \hline 2x - 1 & 6x^2 - 13x + 5 \\ & -(6x^2 - 3x) \\ \hline & -10x + 5 \\ & -(-10x + 5) \\ \hline & 0 \\ \end{array}\) Thus, the quotient is: 3x - 5

Let the universal set be \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) and the set of prime factors of 30 be \(P = \{2, 3, 5\}.\) The complement of \( P \) is \(P' = U - P = \{1, 4, 6, 7, 8, 9\}\) Thus, the complement of \( P \) is {1, 4, 6, 7, 8, 9}.

To find the coordinates of the midpoint \( M \) of the line segment joining the points \( A(3, -4) \) and \( B(-1, 10) \), we use the midpoint formula: \(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\) Substituting the coordinates of points \( A \) and \( B \): - \( x_1 = 3 \) - \( y_1 = -4 \) - \( x_2 = -1 \) - \( y_2 = 10 \) Calculating the coordinates of the midpoint: \(M = \left( \frac{3 + (-1)}{2}, \frac{-4 + 10}{2} \right) = \left( \frac{3 - 1}{2}, \frac{6}{2} \right) = \left( \frac{2}{2}, 3 \right) = (1, 3)\) Thus, the coordinates of the midpoint are (1, 3).

Area of parallelogram PQTU = base * height 32 = 4 * h w = 32/4 w = 8 ∴ = Area of Trapezium PQRU = (1/2)(4 + 14) *8 = 1/2 * (18 * 8) = 72 cm\(^2\)

Mid point = \(\frac{(x_1 + x_2)}{2}\) ; \(\frac{(y_1 + y_2)}{2}\) = \(\frac{(-3 + 5)}{2} ; \frac{(5 - 3)}{2}\) = \(\frac{2}{2}\) ; \(\frac{2}{2}\) = (1, 1)