2010 - JAMB Mathematics Past Questions and Answers - page 3

From the table, the number of students who failed the test is given as:

3 + 1 + 5 = 9

from the table, the number of students who took the test is 2 + 2 + 8 + 4 + 4 = 20

(\begin{array} & Marks(x) & freq.(f) & fx \1 & 2 & 2 \ 2 & 2 & 4 \ 3 & 8 & 24 \ 4 & 4 & 16\ 5 & 4 & 20 \ \hline & \sum f = 20 & \sum fx = 66\end{array})

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Mean mark ,(\bar{x}) = (\frac{\sum fx}{\sum f})

= (\frac{66}{20})

(\bar{x}) = 3.3

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

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})

(\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) = 0 \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}})

Total cost = N34,000 + N2,000

= N36,000

15% = N115

(\frac{115}{100}) x N36,000

= N41,400

W (\alpha) U

W = ku

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

= (\frac{6}{35})

x(x - 1) > 0

x < -1 or x > 1

In the diagram above, < STU = < TRS = 25^o

(< between tangent to circle and a circle and a chord through the point of contact = < in the alternate segment)

< RTS = 90^o

(< in a semicircle = 90^o)

x^o + < RTS + < STU = 180^o

x^o + 90^o + 25^o = 180^o

x^o + 115^o = 180^o

x^o = 180^o - 115^o = 65^o