0.0082N
0.0084N
0.082N
0.084N
Correct answer is B
\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)
\(\therefore F = m(\frac{v - u}{t})\)
\(Mass = 28g = 0.028kg\)
\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)
\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)
\(\begin{pmatrix} 9 \\ -5 \end{pmatrix}\)
\(\begin{pmatrix} -23 \\ -5 \end{pmatrix}\)
\(\begin{pmatrix} 9 \\ 17 \end{pmatrix}\)
\(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)
Correct answer is D
\(\overrightarrow{OY} \equiv -\overrightarrow{YO}\)
Also, \(\overrightarrow{YO} + \overrightarrow{OX} = \overrightarrow{YX}\)
\(\therefore \overrightarrow{YO} = -\overrightarrow{OY} = - \begin{pmatrix} 16 \\ -11 \end{pmatrix} = \begin{pmatrix} -16 \\ 11 \end{pmatrix}\)
\(\overrightarrow{YX} = \begin{pmatrix} -16 \\ 11 \end{pmatrix} + \begin{pmatrix} -7 \\ 6 \end{pmatrix}\)
= \(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)
\(\frac{1}{12}\)
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(\frac{2}{3}\)
Correct answer is D
\(\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}\)
= \(\frac{1}{2} + \frac{1}{6} = \frac{2}{3}\).
Hint: Probability of A and B occurring should be greater than probability A or B happening.
Given that \(a = i - 3j\) and \(b = -2i + 5j\) and \(c = 3i - j\), calculate \(|a - b + c|\).
\(\sqrt{13}\)
\(3\sqrt{13}\)
\(6\sqrt{13}\)
\(9\sqrt{13}\)
Correct answer is B
Given \(a = i - 3j; b = -2i + 5j; c = 3i - j\)
\(a- b + c = (1 - (-2) + 3)i + (-3 - 5 + (-1))j = 6i - 9j\)
\(|a - b + c| = \sqrt{6^{2} + (-9)^{2}} = \sqrt{36 + 81} = \sqrt{117}\)
\(= \sqrt{9 \times 13} = 3\sqrt{13}\)
The marks scored by 4 students in Mathematics and Physics are ranked as shown in the table below
| Mathematics | 3 | 4 | 2 | 1 |
| Physics | 4 | 3 | 1 | 2 |
Calculate the Spearmann's rank correlation coefficient.
0.2
0.5
0.6
0.7
Correct answer is C
| Maths (x) | Rank \(r_{x}\) | Physics (y) | Rank \(r_{y}\) | \(d = |r_{x} - r_{y}|\) | \(d^{2}\) |
| 3 | 2 | 4 | 1 | 1 | 1 |
| 4 | 1 | 3 | 2 | 1 | 1 |
| 2 | 3 | 1 | 4 | 1 | 1 |
| 1 | 4 | 2 | 3 | 1 | 1 |
| Total | 4 |
\(\rho = 1 - \frac{6\sum{d^{2}}}{n(n^{2} - 1)}\)
\( 1 - \frac{6 \times 4}{4(4^{2} - 1)} = 1 - \frac{24}{60}\)
= \(1 - 0.4 = 0.6\)