JAMB Mathematics Past Questions & Answers - Page 310

W \(\alpha\) U

W = ku

u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\)

= \(\frac{6}{35}\)

\(\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 10 \end{array}\)
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\(\bar{x}\) = \(\frac{\sum x }{N}\)

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

= 4

S = \(\sqrt{\frac {(x - \bar{x})^2}{N}}\)

= \(\sqrt{\frac {(10)}{4}}\)

= \(\sqrt{\frac {(5)}{2}}\)

P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)

Using the binomial prob. distribution,

(H + T)3 = H3 + 3H2T1 + 3HT2 + T3

Hence the probability that three heads show in a toss of the three coins is H3

= (\(\frac{1}{2}\))3

= \(\frac{1}{8}\)

A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways

= \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\)

= \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\)

= \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\)

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

= 10 x \(\frac{6 \times 20}{6}\)

= 200

\(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\))

= (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\))

= (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0

= \(\frac{80}{12}

= {\frac {20}{3}}\) or 6\(\frac{2}{3}\)