Explanation

(a) (i) First discount = 20% of N25,000

= \(\frac{20}{100} \times N25,000 = N5,000\)

Further discount for prompt payment = \(2\frac{1}{2}%\) of N25,000.

= \(\frac{5}{200} \times N25,000 = N625\)

\(\therefore \text{Total amount saved by distributor} = N(5,000 + 625)\)

= \(N5625\)

(ii) Total %age discount for prompt payment = \((20 + 2\frac{1}{2})% = 22\frac{1}{2}%\)

Therefore, the distributor paid \((100% - 22\frac{1}{2}%) \times Nx\)

= \(77\frac{1}{2}% \times Nx\)

\(\implies 77\frac{1}{2}% \times x = 11,700\)

\(x = \frac{11,700 \times 200}{155}\)

= \(N15,096.77\)

(b) \(6y^{2} - 149y - 102 \)

= \(6y^{2} - 153y + 4y - 102\)

= \(3y(2y - 51) + 2(2y - 51)\)

= \((3y + 2)(2y - 51)\)

\(\therefore 6y^{2} - 149y - 102 = 0 \implies (3y + 2)(2y - 51) = 0\)

\(3y + 2 = 0 \implies 3y = -2 \)

\(y = -\frac{2}{3}\)

\(2y - 51 = 0 \implies 2y = 51\)

\(y = 25\frac{1}{2}\)

\(\therefore y = -\frac{2}{3} ; y = 25\frac{1}{2}\)

Correct Answer: D. seven

Correct Answer: D. q \(\iff\) \(\sim\)p

Correct Answer: C. 25 - 29

Explanation

The upper quartile occurs at the \(\frac{3N}{4}\) position.

= \(\frac{3 \times 50}{4} = 37.5\)

This occurs in the class 25 - 29.

Explanation

The probability of success (p) = \(\frac{3}{5}\) and the probability of failure (q) = 1 - \(\frac{3}{5} = \frac{2}{5}\)

The probability that at least two of them failed is

P(x \(\geq\) 2) = 1 - p(x < 2) = 1 - [P(x = 0)] + [P(x = 1)]= 1 - [(\(^{10}_{0}\)) (\(\frac{2}{5}\))\(^0\) (\(\frac{3}{5}\))\(^{10}\) + (\(^{10}_1\)) (\(\frac{2}{5}\))\(^1\) (\(\frac{3}{5}\))\(^9\)]

Which simplified to 1 - 0.046357401 = 0.9536 correct to four decimal places

(b) The probability that exactly half of them passes is P(x = 6) = (\(^{10}_{5}\)) (\(\frac{3}{5}\))\(^5\) (\(\frac{2}{5}\))\(^{5}\)

Simplified to give; 252 x 0.7776 x 0.01024 = 0.2007 correct to four decimal places

(c) The probability that at most two of them failed is

P(x \(\leq\) 2) = P(x = 0) + P(x = 1) + P(x = 2). In this case p = \(\frac{2}{5}\) q = \(\frac{3}{5}\) so that

p(x \(\leq\) 2) = (\(^{10}_{0}\)) (\(\frac{2}{5}\))\(^0\) (\(\frac{3}{5}\))\(^{10}\) + (\(^{10}_1\)) (\(\frac{2}{5}\))\(^1\) (\(\frac{3}{5}\))\(^9\) + (\(^{10}_{2}\)) (\(\frac{2}{5}\))\(^2\) (\(\frac{3}{5}\))\(^{8}\)

Which simplies to 0.1673 correct to four decimal places.

Explanation

(a) From y = (x\(^2\) - x + 1)(x - 2) - x + 1)(x - 2)

= x\(^3\)- 3x\(^2\) + 3x - 2

dy/dx = 3x\(^2\) - 6x + 3

=x\(^2\) - 2x + 1

(x - 1) = 0 twice ; x = 1

y = (1\(^2\) - 1 + 1) (1-2) ; y = -1

Gradient m\(_1\) of tangent = y/x = -1/1 = -1

Using m\(_1\)m\(_2\) = -1

Gradient, m\(_2\) of normal =1

Using y-y\(_1\) = m\(_2\)(x - x\(_1\))

y + 1 = x - 1; x - y - 2 = 0

x\(_1\),y\(_1\) = ( -1, 2), , x\(_2\),y\(_2\), =(3, -4)

Mid point of PR =\(\frac{3-1,-4+2}{2,2}\)

= (1-1)

Using \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{y - y_1}{x - x_1}\)

\(\frac{1-{-1}}{5-1}\) = \(\frac{y-1}{x-5}\);

\(\frac{2}{4}\) = \(\frac{y-1}{x-5}\)

2(y-1) = x - 5; 2y - 2 = x - 5

2y - x = -3

x - 2y -3 =0

Correct Answer: D. -1, 0, 1

Explanation

(a) 4 boys, 5 girls

P(both are boys) = P(1st is a boy and 2nd is a boy)

= \(\frac{4}{9} \times \frac{3}{8}\)

= \(\frac{1}{6}\)

(b) If p is the probability of a standard transistors, then \(p = 0.8 = \frac{4}{5}\)

If q is the probability of a transistor below standard, \(q = 0.2 = \frac{1}{5}\)

For 6 transistors, we have

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

(i) p(exactly 2 standard transistors) = \(15p^{2}q^{4}\)

= \(15 \times (\frac{4}{5})^{2} \times (\frac{1}{5})^{4}\)

= \(\frac{48}{3125}\)

(ii) p(exactly 1 standard transistor) = \(6pq^{5}\)

= \(6 \times \frac{4}{5} \times (\frac{1}{5})^{5}\)

= \(\frac{24}{15625}\)

(iii) p(at least 2 standard transistors) = 1 - [p(0 standard transistor) + p(1 standard transistor)]

p(0 standard transistor) = \(q^{6}\)

= \((\frac{1}{5})^{6} = \frac{1}{15625}\)

p(at least 2 standard transistors) = \(1 - [\frac{1}{15625} + \frac{24}{15625}]\)

= \(\frac{15600}{15625} = \frac{624}{625}\)

(iv) p(at most 2 standard transistors) = p(0 standard transistor) + p(1 standard transistor) + p(2 standard transistors)

= \(\frac{1}{15625} + \frac{24}{15625} + \frac{48}{3125}\)

= \(\frac{265}{15625} = \frac{53}{3125}\)

Correct Answer: A. 1 - x

Explanation

If p is the probability of an event happening(occurring) and the probability of an event not occurring is p¹ then p + p¹ = 1

p¹ = 1 - p

x¹ = 1 - x

Correct Answer: (d) 3527