2012 - JAMB Mathematics Past Questions and Answers - page 4

\(\int^2_1(x^2 - 4x)dx\)

\((\frac{x^3}{3} - \frac{4x^2}{2})\)

substituting integrate values

\([\frac{8}{3} - \frac{4 \times 4}{2}] - [\frac{1}{2} - 2]\)

= \(-\frac{11}{3}\)

\(\frac{x}{7} = \frac{96}{1}\) ==> \(\frac{672 + x}{8} = 112\)

Therefore x = 224

Arrange all the values in ascending order,

2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

Range = Highest Number - Lowest Number

Range = 11 - 2 = 9

\(\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \ \hline 2 & -5 & 25 \ \hline 3 & -4 & 16 \ \hline 8 & 1 & 1 \ \hline 10 & 3 & 9 \ \hline 12 & 5 & 25 \ \hline & & 76
\end{array}\)

S.D = \(\sqrt{\frac{(x - \varkappa)^2}{n}}\)

S.D = \(\sqrt{\frac{76}{5}}\)

S.D = 3.9

\(\frac{n + 1 (n - 2)}{(n + 1)!}\)

\(\frac{(n + 1) + (n - 2)!(n - 2)!}{(n + 1)!}\)

\(\frac{(n + 1)(n + 1 -1)(n+1-2)(n+1-3)!}{3!(n-2)!}\)

\(\frac{(n + 1)(n)(n-1)(n-2)!}{3!(n-2)!}\)

\(\frac{(n + 1)(n)(n-1)}{3!}\)

Since n = 15

\(\frac{(15 + 1)(15)(15-1)}{3!}\)

\(\frac{16 \times 15 \times 14}{3 \times 2 \times 1}\)

= 560

\(\frac{8!}{3!}\)

\(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}\)

==> 8 x 7 x 6 x 5 x 4 = 6720

Pass(P) = \(\frac{2}{3}\), Fail(F) = \(\frac{1}{3}\)

T = P.P.F ==> \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}\) = \(\frac{4}{27}\)

Man lives = \(\frac{2}{3}\) not live = \(\frac{1}{3}\)

Wife lives = \(\frac{3}{5}\) not live = \(\frac{2}{5}\)

\(P(\frac{2}{3} \times \frac{2}{5}) + (\frac{2}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{3}{5})\)

= \(\frac{4}{15} + \frac{3}{15} + \frac{6}{15}\)

= \(\frac{13}{15}\)