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\(\frac{8!}{3!}\) \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}\) ==> 8 x 7 x 6 x 5 x 4 = 6720

Pass(P) = \(\frac{2}{3}\), Fail(F) = \(\frac{1}{3}\) T = P.P.F ==> \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\) = \(\frac{4}{27}\)

Man lives = \(\frac{2}{3}\) not live = \(\frac{1}{3}\) Wife lives = \(\frac{3}{5}\) not live = \(\frac{2}{5}\) P(at least one lives to 80 years) = P(man lives to 80 not woman) + P(woman lives to 80 and not man) + P(both live to 80) \(P = (\frac{2}{3} \times \frac{2}{5}) + (\frac{2}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{3}{5})\) = \(\frac{4}{15} + \frac{3}{15} + \frac{6}{15}\) = \(\frac{13}{15}\)

Equation of the line \(\frac{y - 4}{x - 0} = \frac{0 - 4}{1 - 0}\) \(\frac{y - 4}{x} = \frac{-4}{1}\) \(\therefore -4x = y - 4\) \(y = -4x + 4\) \(\therefore \text{The shaded portion = } y \leq -4x + 4\)

< PQR = < PTU = 80^\circ < TSU = 85^\circ x = 180^\circ - (80^\circ + 85^\circ) = 15^\circ

< PQR = 90^\circ (angle in a semi-circle) < QRP = (90 - x)^\circ

Area of trapezium = \(\frac{1}{2} (a + b) \times h\) = \(\frac{1}{2} (7 + 13) \times 6\) = \(60 cm^{2}\)

Excellent = 360^\circ - (120^\circ + 90^\circ + 80^\circ) = 70^\circ Number of excellent students = \(\frac{70}{360} \times 36\) = 7 students

Number that score less than 5: 5 + 6 + 4 + 3 + 2 = 20 Total number of students: 5 + 6 + 4 + 3 + 2 + 4 + 6 + 5 + 1 + 3 + 1 = 40. %age of students that failed: \(\frac{20}{40} \times 100% = 50%\)

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{150}{360} \times \pi \times 12^{2}\) = 60\(\pi\) cm\(^{2}\)

\(\tan 16 = \frac{x}{150}\) \(x = 150 \tan 16\) = \(43 cm\)

If \(y = \cos x\) \(\frac{\mathrm d y}{\mathrm d x}\) =—\(\sin x\)

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

F = mg cos 45o = 1500 x 0.708

S = \(\frac{V^2 - U^2}{20}\) S = \(\frac{0^2 - 30^2}{2 \times 0.1}\)