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})