Correct Answer: C. 15

Correct Answer: C. \(\sqrt{3}\)

Explanation

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

\(\implies \frac{\tan 80 - \tan 20}{1 + \tan 80 \tan 20} = \tan (80 - 20) = \tan 60°\)

\(\tan 60 = \frac{\sin 60}{\cos 60} = \frac{\sqrt{3}}{2} ÷ \frac{1}{2}\)

= \(\sqrt{3}\)

Explanation

(a)

(b) \(\tan \theta = \frac{4.2}{6 + x} ; \tan \theta = \frac{1.2}{x}\)

\(\tan \theta = \tan \theta\)

\(\implies \frac{4.2}{6 + x} = \frac{1.2}{x}\)

\(4.2 x = 1.2(6 + x) \)

\(4.2 x = 7.2 + 1.2 x \implies 4.2x - 1.2x = 7.2\)

\(3x = 7.2 \implies x = 2.4 m\) (length of the shadow cast by the lamp)

(ii) \(\tan \theta = \frac{1.2}{2.4} = 0.5\)

\(\theta = \tan^{-1} (0.5) \)

= \(27°\) (angle of elevation)

Correct Answer: C. 24

Explanation

The median is the middle number of the set, when it is listed in order from smallest to largest or vice versa. In this case we have an odd amount of numbers so we just count from each side until we find the number in the middle. 12,15,16,24,25,31,36 This gives us a final answer of 24 for the median

Explanation

(a)

(b) From the figure above,

(i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.

Correct Answer: B. 10

Explanation

60% boys, 40% girls.

p(boy) = p = 60% = 0.6

p(girl) = q = 40% = 0.4

The binomial distribution function is

\((p + q)^{5} = p^{5} + 5p^{4} q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)

(a) p(exactly 3 boys) = \(10p^{3} q^{2}\)

\(10(\frac{3}{5})^{3} (\frac{2}{5})^{2}\)

= \(10(\frac{27}{125})(\frac{4}{25})\)

= \(\frac{1080}{3125}\)

= \(\frac{216}{625}\)

(b) p(at least 3 are girls) = 1 - [p(0 girls) + p(1 girl) + p(2 girls)].

= \(1 - [p^{5} + 5p^{4}q + 10p^{3} q^{2}]\)

= \(1 - [(\frac{3}{5})^{5} + 5(\frac{3}{5})^{4} (\frac{2}{5}) + 10(\frac{3}{5})^{3} (\frac{2}{5})^{2}]\)

= \(1 - [\frac{243}{3125} + \frac{810}{3125} + \frac{1080}{3125}]\)

= \(1 - [\frac{2133}{3125}]\)

= \(\frac{992}{3125}\).

Explanation

(ii) \(32 - x + x + 28 - x + 12 = 45\)

\(72 - x = 45\)

\(x = 72 - 45 = 27\)

27 students offered both Physics and Government.

(iii) \(n(P \cup G) = 45 - 12 = 33\)

(b) \(p = \frac{2u}{1 - u} ; q = \frac{1 + u}{1 - u}\)

\(\frac{p + q}{p - q}\)

\(p + q = \frac{2u}{1 - u} + \frac{1 + u}{1 - u}\)

= \(\frac{2u + 1 + u}{1 - u}\)

= \(\3u + 1}{1 - u}\)

\(p - q = \frac{2u - 1 - u}{1 - u}\)

= \(\frac{u - 1}{1 - u}\)

= \(\frac{-(1 - u)}{1 - u}\)

= \(-1\)

\(\frac{p + q}{p - q} = \frac{\frac{3u + 1}{1 - u}}{-1}\)

= \(\frac{-(3u + 1)}{1 - u}\)

= \(\frac{-(3u + 1)}{-(u - 1)}\)

= \(\frac{3u + 1}{u - 1}\)

Correct Answer: D. 0.2660

Explanation

(cos 40

Correct Answer: C. (1, -1)

Explanation

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)

\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)

= -1

P = (1, -1)