\(\frac{1}{11}\)
\(\frac{1}{10}\)
\(\frac{1}{6}\)
\(\frac{1}{5}\)
Correct answer is B
Let the number of mixed schools be S
The number of boys only = 42 - S
The number of girls only = 35 - S
(42 - S) + S + (35 - S) = 77 - S = 70
\(\implies\) S = 7
\(Probability = \frac{7}{70} = \frac{1}{10}\)
Evaluate \(\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x\).
\(\frac{8}{3}\)
\(\frac{7}{3}\)
\(\frac{5}{3}\)
2
Correct answer is A
\(\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x \equiv \int_{-1}^{1} (x^{2} + 2x + 1)\mathrm {d} x\)
= \(\left. \frac{x^{3}}{3} + x^{2} + x \right |_{-1}^{1}\)
= \((\frac{1^{3}}{3} + 1^{2} + 1) - (\frac{(-1)^{3}}{3} + (-1)^{2} + (-1)) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)
Calculate the standard deviation of 30, 29, 25, 28, 32 and 24.
2.0
2.8
3.0
3.2
Correct answer is B
| \(x\) | \(x - \mu\) | \((x - \mu)^{2}\) |
| 24 | -4 | 16 |
| 25 | -3 | 9 |
| 28 | 0 | 0 |
| 29 | 1 | 1 |
| 30 | 2 | 4 |
| 32 | 4 | 16 |
| \(\sum\) = 168 | 46 |
\(\mu = \frac{24+25+28+29+30+32}{6} = \frac{168}{8} = 28\)
\(S.D = \sqrt{\frac{\sum{(x - \mu)^{2}}}{n}} = \sqrt{\frac{46}{6}}\)
= \(\sqrt{7.67} \approxeq 2.8\)
Evaluate \(\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x\).
-5.5
-2.0
2.0
5.5
Correct answer is A
\(\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x \equiv \int_{\frac{1}{2}}^{1} 1 - \frac{4}{x^{3}} \mathrm {d} x\).
= \(\int_{\frac{1}{2}}^{1} 1 - 4x^{-3} \mathrm {d} x\).
= \(\left. x - \frac{4x^{-2}}{-2}\right |_{\frac{1}{2}}^{1} = \left. x + \frac{2}{x^{2}}\right |_{\frac{1}{2}}^{1}\)
= \((1 + 2) - (\frac{1}{2} + 8) = 3 - 8.5 = -5.5\)
Find the stationary point of the curve \(y = 3x^{2} - 2x^{3}\).
(1, 0)
(-1, 0)
(1, 1)
(-1, -1)
Correct answer is A
\(y = 3x^{2} - 2x^{3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x - 6x^{2} = 0 \implies 6x(1 - x) = 0\)
\(x = 0, 1\), when x = 1, y = 0.
when x = 1, y = 0.
The stationary point is (1, 0)