JAMB Mathematics Past Questions & Answers - Page 295

$\begin{vmatrix}& \hline 1 & 2 & 3 & 4\\\hline 1 & 1 & 2& 3 & 4\\2 & 2& 4 & 6 & 8\\ 3& 3& 6& 9 & 12\\4 & 4 & 8 & 12& 16\end{vmatrix}$

p (product of x, and y $\leq$ 6) = $\frac{10}{16}$

= $\frac{5}{8}$

p(x) = $\frac{2}{5}$ p(y) = $\frac{1}{3}$

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= $\frac{2}{5}$ + $\frac{1}{3}$

= $\frac{11}{5}$

Number of red balls = 16,

Number of blue balls = 20

Let x represent the No of white balls to be added

∴ Total number of balls = 36 + x

2(36 + x) = 80

= 2x + 80 - 72

= 8

x = $\frac{8}{2}$

= 4

mean (x) = $\frac{1 + x + 1 + 2x + 1}{3}$

= $\frac{3x + 3}{3}$

= 1 + x

$\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline 1 & -x & x^2 \\ x + 1 & 0 & 0\\2x + 1 & x & x^2\\ \hline & & 2x^2\end{array}$

S.D = $\sqrt{\frac{\sum(x - 7)^2}{\sum f}}$

= $\sqrt{(6)}^2$

= $\frac{2x^2}{3}$

= 2x2

= 18

x2 = 9

∴ x = $\pm$ $\sqrt{9}$

= $\pm$3

mean (x) = $\frac{\sum x}{N}$

= k + k + 1 + k + 3

= $\frac{3k + 3}{3}$

= k + 1

$\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline k & -1 & 1 \\ k + 1 & 0 & 0\\ k + 2 & 1 \\ \hline & & 2\end{array}$

Variance (52) = $\frac{\sum (x - x)^2}{N}$

= $\frac{2}{3}$