Explanation

Note that P(X not hitting the target) = \(\frac{2}{3}\)

P(Y not hittiing the target) = \(\frac{4}{5}\) and P(Z not hitting the target) = \(\frac{3}{4}\)

So that, P(only one) = (\(\frac{1}{3} \times \frac{4}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{1}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{4}{5} \times \frac{1}{4}\)) = \(\frac{12}{60} + \frac{6}{60} = \frac{13}{30}\)

= 0.43, correct to two decimal places.

Correct Answer: A. perpendicular bisector of the straight line joining them

Correct Answer: D. 95°

Explanation

In the diagram,

a = 50\(^o\) (alternate angles)

b\(_1\) + a 35\(^o\) = 180\(^o\) (sum of angles on a straight line)

i.e; b\(_1\) + 50\(^o\) + 35\(^o\)

= 180v

b\(_1\) + 180\(^o\) - 85\(^o\) = 90\(^o\)

But b\(_2\) = \(b_1\) = 95\(^o\) (angles in alternate segement)

<QST = b\(_2\) = 95\(^o\)

Correct Answer: (c) 12

Correct Answer: B. \(3\sqrt{13}\)

Explanation

Given \(a = i - 3j; b = -2i + 5j; c = 3i - j\)

\(a- b + c = (1 - (-2) + 3)i + (-3 - 5 + (-1))j = 6i - 9j\)

\(|a - b + c| = \sqrt{6^{2} + (-9)^{2}} = \sqrt{36 + 81} = \sqrt{117}\)

\(= \sqrt{9 \times 13} = 3\sqrt{13}\)

Correct Answer: D. 55°

Explanation

50\(^o\) + m = 105\(^o\) (exterior angle of \(\triangle\)

m = 105\(^o\) - 50\(^o\)

m = 55\(^o\)

Correct Answer: D. 6N

Explanation

F = m * a

d = t\(^2\) + 3t.

a = \(\frac{d^2d}{dt^2}\)

\(\frac{d[d]}{dt}\) = 2t + 3

\(\frac{d^2d}{dt^2}\) = 2m/s\(^2\)

a = 2m/s\(^2\)

F = m * a

F = 3 × 2 = 6N

Correct Answer: D. \(\frac{1}{6}(10\sqrt{3}-9\sqrt{2}\)

Explanation

\(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\\

=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\\

\frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\\

\frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\\

\frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)

Explanation

(a)(i)

(ii) \(21 + 4 + 5 + 10 + x + 35 - x + 32 - x = 100\)

\(107 - x = 100\)

\(x = 107 - 100 = 7\)

Therefore, 7 traders sell oranges and mangoes alone.

(b) \(312_{four} + 52_{x} = 96_{ten}\)

\(312_{4} = 3 \times 4^{2} + 1 \times 4^{1} + 2 \times 4^{0}\)

= \(48 + 4 + 2 = 54_{10}\)

\(52_{x} = 5 \times x^{1} + 2 \times x^{0}\)

= \((5x + 2)_{10}\)

\(\therefore 54 + (5x + 2) = 96\)

\(56 + 5x = 96 \implies 5x = 96 - 56 = 40\)

\(x = 8\)

\(\therefore 312_{4} + 52_{8} = 96_{10}\)

Explanation

(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)

= \(\frac{3150}{40}\)

= 78.75

(b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\)

= \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\)

= \(\sqrt{6264.75 - (78.75)^{2}}\)

= \(\sqrt{6264.75 - 6201.5625}\)

= \(\sqrt{63.1875}\)

= 7.95.