Further Mathematics Questions and Answers

No explanation has been provided for this answer.

\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)

\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)

\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)

\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)

Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)

\(\frac{(n - 2)(n - 3)!}{(n - 3)! 3!} = 1\)

\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)

\(n = 2 + 6 = 8\)

Given line: \(3x - 2y + 4 = 0 \implies 2y = 3x + 4\)

\(y = \frac{3}{2}x + 2\)

\(Gradient (\frac{\mathrm d y}{\mathrm d x}) = \frac{3}{2}\)

Gradient of perpendicular line = \(\frac{-1}{\frac{3}{2}} = \frac{-2}{3}\)

\(\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}\)

\(\frac{y + 3}{x - 2} = \frac{-2}{3} \)

\(3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0\)

= \(3y + 2x + 5 = 0\)

 \(V = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\) and \(U = \begin{pmatrix} -1 \\ 5 \end{pmatrix}\)

\(U + V = \begin{pmatrix} -1 - 2 \\ 5 + 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}\)

\(|U + V| = \sqrt{(-3)^{2} + 9^{2}} = \sqrt{9 + 81} = \sqrt{90}\)

= \(sqrt{9 \times 10} = 3\sqrt{10}\)

Mean (\(\bar{x}\)) = \(\frac{1+2+3+4+5+5+6+7+8+9}{10} = \frac{50}{10} = 5\)

\(MD = \frac{\sum |x - \bar{x}|}{n} = \frac{20}{10} = 2\)