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Arrange all the values in ascending order, 2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

Range = Highest Number - Lowest Number Range = 11 - 2 = 9

\(\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \\ \hline 2 & -5 & 25 \\ \hline 3 & -4 & 16 \\ \hline 8 & 1 & 1 \\ \hline 10 & 3 & 9 \\ \hline 12 & 5 & 25 \\ \hline & & 76 \end{array}\) S.D = \(\sqrt{\frac{(x - \varkappa)^2}{n}}\) S.D = \(\sqrt{\frac{76}{5}}\) S.D = 3.9

\(\frac{n + 1 (n - 2)}{(n + 1)!}\) \(\frac{(n + 1) + (n - 2)!(n - 2)!}{(n + 1)!}\) \(\frac{(n + 1)(n + 1 -1)(n+1-2)(n+1-3)!}{3!(n-2)!}\) \(\frac{(n + 1)(n)(n-1)(n-2)!}{3!(n-2)!}\) \(\frac{(n + 1)(n)(n-1)}{3!}\) Since n = 15 \(\frac{(15 + 1)(15)(15-1)}{3!}\) \(\frac{16 \times 15 \times 14}{3 \times 2 \times 1}\) = 560

\(\frac{8!}{3!}\) \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}\) ==> 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: \(\frac{20}{40} \times 100% = 50%\)

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

Divide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = \(\frac{1050}{70}\) = 15 and y = \(\frac{1190}{70}\) = 17 ratio of enrollment of x to y = 15:17