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 \(\sin \theta = +ve\).

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

Correct Answer: B. {1, 2, 3, 4, 6}

Explanation

\(\mu\) = {1, 2, 3, 4..., 10}

X = {1, 2, 3, 4, 5, 6}

Y = {1, 2, 3, 4, 6, 8}

Therefore;

X \(\cap\) Y = {1, 2, 3, 4, 6}

Explanation

(a) 1st bag contains 12 white (W) balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.

P(two black balls from the 1st bag and two black balls from 2nd bag)

= \((\frac{8}{20} \times \frac{7}{19}) \times (\frac{15}{25} \times \frac{14}{24})\)

= \(\frac{14}{95} \times \frac{7}{20}\)

= \(\frac{49}{950}\)

(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)

= \(\frac{12}{20} \times \frac{8}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{12}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{10}{25} \times \frac{15}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{15}{25} \times \frac{10}{24}\)

= \(\frac{42}{475} + \frac{42}{475} + \frac{7}{190} + \frac{7}{190}\)

= \(\frac{238}{950}\)

Correct Answer: D. 16/9

Explanation

P ? 1/w²; P = K/W²

9 = K/4²; 9 = K/16

K = 9 x 16 = 144

P = K/W²; when w = 9

P = 144/9² = 144/81 = 16/9

Correct Answer: A. 4.0 x 10^-3

Correct Answer: C. 1/2

Explanation

(a)

(b) Let the assumed mean be 25.5 yrs.

Mean = \(A + \frac{\sum fd}{\sum f}\)

= \(25.5 + \frac{265.5}{63}\)

= \(25.5 + 4.21\)

= \(29.71 yrs\)

Correct Answer: D. 11

Correct Answer: D. 3x² - 11x + 6 = 0

Explanation

x = 3; x = \(\frac{2}{3}\).

(x - 3)(x - \(\frac{2}{3}\)) = 0

\(x^2 - \frac{2x}{3} - 3x + 2 = 0\)

\(x^2 - \frac{11x}{3} + 2 = 0\)

\(3x^2 - 11x + 6 = 0\)

Correct Answer: A. \(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\)

Explanation

\(f(x) = \frac{2x - 1}{x + 2}\)

\(y = \frac{2x - 1}{x + 2}\)

\(x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1\)

\(xy - 2y = -1 - 2x \implies y = \frac{-1 - 2x}{x - 2}\)

\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)