Correct Answer: B. 0.22

Explanation

The probability of picking a black ball = \(\frac{6}{6 + 14} = \frac{6}{20}\)

Probability of a white ball without replacement = \(\frac{14}{19}\)

Probability of a black ball and then white without replacement = \(\frac{6}{20} \times \frac{14}{19} = \frac{21}{95} = 0.22\)

Explanation

f(x) =2xp\(^3\) + px\(^2\) + qx - 5

f(1) = 2(1)\(^3\) + p(1)\(^2\) + q(1) - 5

2+ p + q - 5;

p + q =3 -- (i)

f(-2) = 27

i.e, 2(-2)\(^3\) + p(-2)\(^2\) +q(-2) - 5 = 27

-16 +4p - 2q - 5 = 27

2p - q = 24--- (ii)

Solving(i) and (ii) Simultaneously

P = 9, q= -6

Explanation

(a) 2M, 5P, 3C.

(i) Tie same books together

Maths books can be arranged within themselves in 2! ways, Physics books can be arranged in 5! ways, Chemistry books in 3! ways.

The three sets of subjects can be arranged in 3! ways.

\(\therefore\) Total arrangements = 3!2!5!3!

= \(6 \times 2 \times 120 \times 6 = \text{8640 ways}\).

(ii) Tie Physics books together.

There are 1 + 2 + 3 = 6 books to be arranged in 6! ways.

The Physics books can be arranged in 5! ways.

\(\therefore\) Total arrangement = \(6!5! = 720 \times 120 = \text{86,400 ways}\)

(b) p(English speaking) = p = \(\frac{13}{20}\); p(non- English speaking) = q = \(\frac{7}{20}\)

8 persons selected,

\((p + q)^{8} = p^{8} + 8p^{7}q + 28p^{6}q^{2} + 56p^{5}q^{3} + 70p^{4}q^{4} + 56p^{3}q^{5} + 28p^{2}q^{6} + 8pq^{7} + q^{8}\)

p(at least 3 speak English) = 1 - [none + one + two].

= \(1 - [q^{8} + 8pq^{7} + 28p^{2}q^{6}]\)

= \(1 - [(0.35)^{8} + 8(0.65)(0.35)^{2} + 28(0.65)^{2}(0.35)^{2}]\)

= \(1 - [0.0002252 + 0.003346 + 0.1131]\)

= \(1 - 0.001488 = 0.98512\)

\(\approxeq 0.985\) (3 sig. figs).

Correct Answer: B. 1

Explanation

x? \(=\frac{∑fx}{∑f}= 3\)

\(=\frac{(1 \times 2)+(2 \times k)+(3 \times 1)+(4 \times 1)+(5 \times 2)}{2 + k + 1 + 1 + 2}= 3\)

\(=\frac{2 + 2k + 3 + 4 + 10}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = \frac{3}{1}\)

=19+2k=3(6+k)

=19+2k=18+3k

=2k-3k=18-19

=-k=-1

∴k=1

Explanation

Observe that the common difference = 2, substitute to get 165 = \(\frac{n}{2}\)[2(-3) + (n - 1)2]

n\(^2\) - 4n - 165 = 0

n = 15 or n = -11

Therefore, the number of terms is 15 as n cannot be negative.

Correct Answer: C. 6006

Explanation

No of boys in the class = 7; Girls = 20-7 = 13

No of selection = \(^{13}C_{3} \times ^{7}C_{2} = \frac{13!}{(13-3)!3!} \times \frac{7!}{(7-2)!2!}\)

= \(286\times21 = 6006\)

Correct Answer: D. 4

Correct Answer: A. 300°

Explanation

In order to do this, simply find the option in the range where only the cos is +ve. This occurs in the range \(270° \leq x \leq 360°\).

Check: \(\sin 300 = - \sin 60 = \frac{-\sqrt{3}}{2}\)

\(\cos 300 = \cos 60 = \frac{1}{2}\)

Explanation

\(a \Delta b = a^{3} - b^{3}\)

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

\((\sqrt{3} + \sqrt{2})^{3} = (\sqrt{3})^{3} + 3(\sqrt{3})^{2}(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^{2} + (\sqrt{2})^{3}\)

= \(3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}\)

= \(9\sqrt{3} + 11\sqrt{2}\)

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

= \(3\sqrt{3} - 9\sqrt{2} + 6\sqrt{3} - 2\sqrt{2}\)

= \(9\sqrt{3} - 11\sqrt{2}\)

\((\sqrt{3} + \sqrt{2}) \Delta (\sqrt{3} - \sqrt{2}) = 9\sqrt{3} + 11\sqrt{2} - 9\sqrt{3} + 11\sqrt{3}\)

= \(22\sqrt{2}\)

Correct Answer: D. {36}

Explanation

P = {3,6,9,12,18,36}; Q = { 4,36}

P n Q = {36}