(12N, 090°)
(10N, 270°)
(10N, 180°)
(10N, 120°)
Correct answer is B
No explanation has been provided for this answer.
If \(\frac{^{n}C_{3}}{^{n}P_{2}} = 1\), find the value of n.
8
7
6
5
Correct answer is A
\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)
\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)
\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)
\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)
Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)
\(\frac{(n - 2)(n - 3)!}{(n - 3)! 3!} = 1\)
\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)
\(n = 2 + 6 = 8\)
2y - 3x = 0
3y - 2x + 5 = 0
3y + 2x + 5 = 0
2y - 3x - 5 = 0
Correct answer is C
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 = \(\frac{-1}{\frac{3}{2}} = \frac{-2}{3}\)
\(\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}\)
\(\frac{y + 3}{x - 2} = \frac{-2}{3} \)
\(3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0\)
= \(3y + 2x + 5 = 0\)
\(3\sqrt{10}\)
\(\sqrt{82}\)
15
\(2\sqrt{5}\)
Correct answer is A
\(V = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\) and \(U = \begin{pmatrix} -1 \\ 5 \end{pmatrix}\)
\(U + V = \begin{pmatrix} -1 - 2 \\ 5 + 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}\)
\(|U + V| = \sqrt{(-3)^{2} + 9^{2}} = \sqrt{9 + 81} = \sqrt{90}\)
= \(sqrt{9 \times 10} = 3\sqrt{10}\)
Calculate the mean deviation of 1, 2, 3, 4, 5, 5, 6, 7, 8, 9.
2
3
4
5
Correct answer is A
| x | 1 | 2 | 3 | 4 | 5 | 5 | 6 | 7 | 8 | 9 | Total |
| \(x - \bar{x}\) | -4 | -3 | -2 | -1 | 0 | 0 | 1 | 2 | 3 | 4 | |
| \(|x - \bar{x}|\) | 4 | 3 | 2 | 1 | 0 | 0 | 1 | 2 | 3 | 4 | 20 |
Mean (\(\bar{x}\)) = \(\frac{1+2+3+4+5+5+6+7+8+9}{10} = \frac{50}{10} = 5\)
\(MD = \frac{\sum |x - \bar{x}|}{n} = \frac{20}{10} = 2\)