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(x) = \(\frac{5 + 6 + 7}{3}\) = \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\\hline 5 & 5 - 6 & 1\\ 6 & 6 - 6 & 0 \\ 7 & 7 - 6 & 1\\ \hline & & 2\end{array}\) S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x) = \(\sqrt{\frac{2}{3}}\)

\(\begin{array}{c|c} & & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a & 1 & 1, 1 & 1, 2 & 1, 3 & 1, 4 & 1, 5 & 1, 6\\\hline & 2 & 1,1 & 2, 2 & 2, 3 & 2, 4 & 2, 5 & 2, 6 \\\hline B & 3 & 3, 1 & 3, 2 & 3, 3 & 3, 4 & 3, 5 & 3, 6\\\hline & 4 & 4 , 1 & 4, 2 & 4, 3 & 4, 4 & 4, 5 & 4, 6\\\hline Die & 5 & 5, 1 & 5, 2 & 5, 3 & 5, 4 & 5, 5 & 5, 6 \\\hline & 6 & 6, 1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6, 6\end{array}\) Probability of obtaining total score of 8 = \(\frac{5}{36}\)

Prob of P = \(\frac{3}{4}\) Prob of Q = \(\frac{1}{6}\) Prob of both P or Q = \(\frac{3}{4} + \frac{1}{6} = \frac{11}{12}\)

Treat the couple as one block, giving us 4 units to arrange (the couple + 3 others). Arrange the 4 units: 4! = 4 x 3 x 2 x 1 = 24. Arrange the couple within their block: 2! = 2 x 1 = 2. Total arrangements: 4! \times 2! = 24 \times 2 = 24 \times 2 = 48. Thus, the total number of arrangements is 48.

Total Mathematics courses = 4 Total Physics courses = 3 Courses to choose = 5 Courses in Physics to choose = 2 Courses in Mathematics to choose = 3 Mathematics = \(^{4} C_{3} = 4\) Physics = \(^{3} C_{2} = 3\) \(\therefore\) He can choose in 4 x 3 = 12 ways.

Given the graph and the curve y = ax2 + bx + c the roots are x - 2 and 2 while its equation (x + 2)(x - 2) = y y = x2 - 4 i.e. y = x2 + 0x - 4 a = 1, b = 0 and c = -4

PSR = 80^\circ \(\therefore\) SRP = 20^\circ \(\implies\) RPQ = 20^\circ x = 140^\circ

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

Area of the quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (6)^2\) = \(\frac{36 \pi}{4} = 9 \pi \) Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\) Area of shaded portion =(9 \(\pi \) - 18)cm^2\) = 9(\(\pi - 2)cm^2\).

RY = YZ - RZ = 10 tan 45 - 10 tan 30 = 10(1 - \(\frac{1}{\sqrt{3}}\))

(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\) \((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\) \(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\) \(3^{4x - 2} = 3^{4x - 8 - x}\) \(3^{4x - 2} = 3^{3x - 8}\) \(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\) \(x = -6\) (b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\) = \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\) = \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\) = \(0.6 \times 10^{-6}\) = \(6 \times 10^{-7}\)

(a) \(2x + 5y = \frac{13}{2} \implies 4x + 10y = 13 ... (1)\) \(5x - 2y = 9 ... (2)\) \((2) \times 5 : 25x - 10y = 45 ... (2a)\) \((1) + (2a) : 29x = 58 \implies x = 2\) \(4(2) + 10y = 13 \implies 10y = 13 - 8 = 5\) \(y = \frac{5}{10} = 0.5\) (b) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\) \(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\) \(\frac{2x + 1}{3x - 2} = 10^{1} = 10\) \(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\) \(21 = 28x \implies x = \frac{3}{4}\)

(a) Length of arc AB = \(\frac{108^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{132}{10} =13.2cm\) Perimeter of the sector = \(13.2cm + 7cm + 7cm = 27.2cm\) (b) Let T = top of cliff, C = bottom of cliff, B = the boat. Since angle of depression = 30^\circ then < CBT = 30^\circ (alternate angles TR//CB) \(\therefore \text{Height of the cliff} (CT) = 120 \times 0.5774m\) = \(69.288m \approxeq 69.3m\)

(a) < PQR = 180^\circ - 131^\circ = 49^\circ (angles on a straight line) \(\therefore < PQR = 49^\circ\) Also, \(< PQR = 180^\circ - 98^\circ = 82^\circ\) \(\therefore < QPR = 180^\circ - (49^\circ + 82^\circ) = 180^\circ - 131^\circ\) = \(49^\circ\) (b) Circumference of the circular track = \(2\pi r\) \(\therefore 2\pi r = 400 \) where \(\pi = \frac{22}{7}\) \(\therefore r = \frac{400 \times 1 \times 7}{22 \times 2} = \frac{700}{11}\) = \(63.63m \approxeq 64m\)

(a) P(obtaining a 3 from a die) = \(\frac{1}{6}\) P(obtaining a tail in 2 coins tossed at once) = \(\frac{1}{2}\). \(\therefore\) P(obtaining 3 and a tail = \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\) (b) No of black marbles = n(B) = 7 No of red marbles = n(R) = 3 Total marble = 10 (i) P(red then black) = P(1st red, then black) or P(1st black, then red) = \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\) = \(\frac{7}{15}\) (ii) Prob (two blacks) = \(\frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\)