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at most 2 marks = 5 + 2 + 4 students = 11 students probability(at most 2 marks) = \(\frac{11}{20}\)

\(\sqrt{12} ( \sqrt{48} - \sqrt{3}\)) \(\sqrt{4 \times 3} (6 \times 3 - \sqrt{3}) = 2 \sqrt{3}(4 \sqrt{3} - \sqrt{3})\) = 2\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\)

x > y and 3 3 x > 3 to give the possible x > y > 3

let the number be x 2\(\frac{1}{2}x + \frac{3}{4}x = 13\) \(\frac{5}{2}x + \frac{3}{4}x = 13\) multiply through by 4 4(\(\frac{5}{2}\))x + 4(\(\frac{3}{4}\))x = 13 x 4 2(5x) + 3x = 52 10x + 3x = 52 13x = 52 x = \(\frac{52}{13}\) x = 4

x = {0, 2, 4, 6}; y = {1, 2, 3, 4}; z = {1, 3} u = {0, 1, 2, 3, 4, 5, 6} y' = {0, 5, 6} to find x \(\cap\) (Y' \(\cup\) Z) first find y' \(\cup\) z = {0, 1, 3, 5, 6} then x \(\cap\) (Y' \(\cup\) Z) = {0, 6}

x2 = 3(2x + 9) x2 = 6x + 27 x2 - 6x - 27 = 0 x2 - 9x + 3x - 27 = 0 x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0 x + 3 = 0 or x - 9 = 0 x = -3 or x = 9 x = -3, x = 9

p(4, 3) Q(2 - 1) distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\) = \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\) = \(\sqrt{(-2)^2 = (-4)^2}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\)

E = \(\frac{m}{2g}\)(v2 - u2) multiply both sides by 2g 2Eg = 2g (\(\frac{M}{2g} (V^2 - U^2)\) 2Eg = m(V2 - U2) 2Eg - mV2 - mU2 mU2 = mV2 - 2Eg divide both sides by m \(\frac{mU^2}{m} = \frac{mV^2 - 2Eg}{m}\) U2 = \(\frac{mV^2 - 2Eg}{m}\) = \(\frac{mV^2}{m} - \frac{2Eg}{m}\) U2 = V2 - \(\frac{2Eg}{m}\) U = \(\sqrt{V^2 - \frac{2Eg}{m}}\)

x \(\alpha \frac{1}{y}\) y \(\alpha z\) the relationship = x \(\alph \frac{1}{z}\)

(2, -3)(2, 5) gradient = \(\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2}\) = \(\frac{5 + 3}{0}\) = \(\frac{8}{0}\) = undefined

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

\(\begin{array}{c|c} height(x) & frequency(f) & fx \\ \hline 2 & 2 & 4\\ 3 & 4 & 12\\ 4 & 5 & 20 \\ 5 & 3 & 15\\ 6 & 1 & 6\end{array}\) mean\(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{57}{15}\) = 3.8

diameter = 12cm radius = \(\frac{12}{2}\)cm = 6cm area of sector = 66cm2 area of sector = area of cone 66 = \(\pi rl\) 66 = \(\frac{22}{7} \times r \times 6\) divide both sides by 6 11 = \(\frac{22}{7}r\) 22r = 11 x 7 r = \(\frac{77}{22}\) = 3.5cm

let the chord be AB = 7cm radius OA = 3.7cm distance of the = OM using Pythagoras theorem OA2 = AM2 + OM2 3.72 = 3.52 + OM2 13.69 = 12.25 + OM2 13.69 - 12.25 = OM2 1.44 = OM2 OM = \(\sqrt{1.44}\) OM = 1.2cm distance of chord = 1.2cm

The range is a measure of dispersion that indicates the difference between the highest and lowest values in a dataset.