Correct Answer: C. -6

Explanation

\(4p_2 + 4C_2 - 4p_3\)

\(np_r = \frac{n!}{[n-r]!} and nC_r = \frac{n!}{[n-r]!r!} \)

= \(\frac{4!}{[4-2]!} + \frac{4!}{[4-2]!2!} - \frac{4!}{[4-3]!} = \frac{4!}{2!} + \frac{4!}{2!2!} - \frac{4!}{1!}\)

= \(\frac{4*3*2!}{2!} + \frac{4*3*2!}{2!2!} - \frac{4*3*2*1}{1!}\)

12 + 6 - 24 = -6

Correct Answer: D. \(\frac{-4}{5}\)

Explanation

Given \(\sin \theta = \frac{3}{5} \implies opp = 3, hyp = 5\)

Using Pythagoras' Theorem, we have \( adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16} = 4\)

\(\therefore \cos \theta = \frac{4}{5}, 0° < \theta < 90°\)

In the quadrant where \(180° - \theta\) lies is the 2nd quadrant and here, only \(\sin \theta = +ve\).

\(\therefore \cos (180 - \theta) = -ve = \frac{-4}{5}\)

Explanation

p(have valid license) = p = 0.8

p(no valid license) = q = 0.2.

The binomial probability distribution is

\((p + q)^{6} = p^{6} + 6p^{5} q + 15p^{4} q^{2} + 20p^{3} q^{3} + 15p^{2} q^{4} + 6p q^{5} + q^{6}\)

(a) p(2 have valid licenses) = \(15p^{2} q^{4}\)

= \(15 (0.8)^{2} (0.2)^{4} = 15(0.64)(0.0016)\)

= \(0.01536 \approxeq 0.015\) (3 d.p)

(b) p(more than 3 have valid licenses) = \(p^{6} + 6p^{5} q + 15p^{4} q^{2}\)

= \((0.8)^{6} + 6(0.8)^{5} (0.2) + 15(0.8)^{4} (0.2)^{2}\)

= \(0.262144 + 0.393216 + 0.24578\)

= \(0.90112 \approxeq 0.901\) (3 d.p)

(c) p(at least 5 have valid licenses) = \(p^{6} + 6p^{5} q\)

= \((0.8)^{6} + 6(0.8)^{5} (0.2)\)

= \(0.262144 + 0.393216\)

= \(0.65536 \approxeq 0.655\).

Correct Answer: C. 31.5m

Explanation

S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);

ds/dt = 12 + 5t - 3t\(^2\)

At max height ds/dt = 0

i.e 12 + 5t - 3t\(^2\)

(3t + 4)(t -3) = 0;

t = -4/3 or 3

Hmax = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)

= 36 + 45/2 - 27

= 31.5m

Correct Answer: A.\(\frac{-x}{1 - x}, x \neq 1\)

Explanation

\(x * y = x + y - xy\)

Let \(x^{-1}\) be the inverse of x, so that

\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)

\(x + x^{-1} - x(x^{-1}) = 0 \implies x(x^{-1}) - x^{-1} = x\)

\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)

= \(\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1\)

Correct Answer: D. 1 + \(\sqrt{6}\)

Explanation

x*y = x\(^2\) - y\(^2\) + xy

(\(\sqrt{3}\))*(\(\sqrt{2}\)) = (\(\sqrt{3}\))\(^2\) - (\(\sqrt{2}\))\(^2\) + \(\sqrt{3}\) x \(\sqrt{2}\)

= 3 - 2 + \(\sqrt{6}\)

= 1 + \(\sqrt{6}\)

Correct Answer: B. 75N

Explanation

Let the normal reaction be R

R = Wcosø = mgcosø

= 15 * 10cos 60°

= 150 * 0.5000

= 75N

Explanation

Resultant :

\(R = R_{1} + R_{2} + R_{3}\)

= \(\begin{pmatrix} -4 \cos 45 \\ -4 \sin 45 \end{pmatrix} + \begin{pmatrix} 0 \\ 3 \sin 90 \end{pmatrix} + \begin{pmatrix} 2 \cos 45 \\ 2 \sin 45 \end{pmatrix}\)

= \(\begin{pmatrix} -2\sqrt{2} \\ -2\sqrt{2} \end{pmatrix} + \begin{pmatrix} 0 \\ 3 \end{pmatrix} + \begin{pmatrix} \sqrt{2} \\ \sqrt{2} \end{pmatrix}\)

= \(\begin{pmatrix} - \sqrt{2} \\ 3 - \sqrt{2} \end{pmatrix}\)

\(|R| = \sqrt{(-\sqrt{2})^{2} + (3 - \sqrt{2})^{2}}\)

= \(\sqrt{2 + 11 - 2\sqrt{2}}\)

= \(\sqrt{13 - 2\sqrt{2}}\)

= \(10.1718\)

\(|R| = 3.1893 \approxeq 3.2N\)

Correct Answer: D. \(y = 3x^{2} - 2x + 4\)

Explanation

The criterion for the quadratic curve to intersect the x- axis is \(b^{2} > 4ac\).

Explanation

(a) \(F = xi + 15j\)

\(|F| = \sqrt{x^{2} + 15^{2}} = 17\)

\(x^{2} = 17^{2} - 15^{2} = 64\)

\(x = \pm \sqrt{64} = \pm 8\)

\(x = 8 ; x = -8\)

(b)

\(\tan \alpha = \frac{15}{8} = 1.875\)

\(\alpha = \tan^{-1} (1.875) = 61.9° \approxeq 62°\)

\(\tan \beta = \frac{15}{-8} = -1.875\)

\(\beta = \tan^{-1} (-1.875) = -61.9°\)

= \(180° - 61.9° = 118.1° \approxeq 118°\)

Directions of forces are 62° and 118° each with the positive x- axis.