JAMB Mathematics Past Questions & Answers - Page 228

The total sample space when two dice are thrown together is 6 x 6 = 36

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\)

At least 10 means 10 and above

P(at least 10) = \(\frac{6}{36}\)

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

Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6

Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)

\(\frac{31}{2}\) = 15.5

\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065

1.0(1 d.p)

\(\bigtriangleup\)PUS is right angled

\(\frac{US}{5}\) = sin60^o

US = 5 x \(\frac{\sqrt{3}}{2}\)

= 2.5\(\sqrt{3}\)

= 4.33cm

The volume of the pipe is equal to the area of the cross section and length.

let outer and inner radii be R and r respectively.

Area of the cross section = (R² - r²)

where R = 6 and r = 6 - 1

= 5cm

Area of the cross section = (6² - 5²)\(\pi\) = (36 - 25)\(\pi\)cm sq

vol. of the pipe = \(\pi\) (R² - r²)L where length (L) = 10

volume = 11\(\pi\) x 10

= 110\(\pi\)cm³

\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)

from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii)

Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \)

= \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\)

\(2\cos^2 \theta = \frac{7}{8}\)

\(\cos^2 \theta = \frac{7}{2 \times 8}\)

\(\frac{7}{16} = \cos \theta\)

\(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)

but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)

\(opp^2 = hyp^2 - adj^2\)

\(opp^2 = 4^2  - (\sqrt{7})^{2}\)

= 16 - 7

opp = \(\sqrt{9}\) = 3

\(\tan \theta = \frac{\text{opp}}{\text{adj}}\)

= \(\frac{3}{\sqrt{7}}\)

\(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)