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\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\) mode = 9.5 + \(\frac{D_1}{D_2 + D_1}\) x C = 9.5 + \(\frac{5(32 - 17)}{2(32) - 17 - 25}\) = 9.5 + \(\frac{75}{22}\) = 12.91 \(\approx\) 12.9

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) = \(\frac{\sum (x - x)^2}{N}\) = \(\frac{2}{3}\)

mean (x̄) = \(\frac{1 + x + 1 + 2x + 1}{3}\) = \(\frac{3x + 3}{3}\) = 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 = \(\sqrt{\frac{\sum(x - x)^2}{n}}\) = \(\sqrt{(6)}\) ( given ) \(\sqrt{(6)}\) = \(\sqrt{\frac{2x^2}{3}}\) cancel square roots and cross multiply \(2x^2\) = 18 divide both sides by 2 \(x^2\) = 9 ∴ x = \(\pm\) \(\sqrt{9}\) = \(\pm\)3

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 Pr( red ball) = \(\frac{ 16}{ 36 + x } = \frac{2}{5}\) (remember that Pr( red ball ) = \frac{2}{5}\)) cross multiplying 2(36 + x) = 80 2x = 80 - 72 2x = 8 x = \(\frac{8}{2}\) = 4

To find the probability that either team X or team Y wins the gold, we can use the formula for the probability of the union of two events: P(x or y) = p(x) + p(y) - p(x n y) But p(x n y ) = 0 p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\) p(x or y) = p(x ∪ y) = p(x) + p(y) = \(\frac{2}{5}\) + \(\frac{1}{3}\) = \(\frac{11}{5}\)

\(\begin{vmatrix}& \hline 1 & 2 & 3 & 4\\\hline 1 & 1 & 2& 3 & 4\\2 & 2& 4 & 6 & 8\\ 3& 3& 6& 9 & 12\\4 & 4 & 8 & 12& 16\end{vmatrix}\) p (product of x, and y \(\leq\) 6) = \(\frac{10}{16}\) = \(\frac{5}{8}\)

y = 6 sin (2x - 25)º \(\frac{dy}{dx}\) = 12 cos(2x - 25)º \(\frac{dy}{dx}\) = 0 12 cos x (2x - 25º) = 0 = cos(2x - 25º) = 0 = 2x - 25º = cos\(^{-1}\)(0) = 2x - 25º = 90º x = \(\frac{115}{2}\)

m = \(\frac{y_2 - y_1}{x_2 - x_2} = \frac{3 - 0}{0 - 2} = \frac{-3}{2}\) = \(\frac{y - y_1}{x- x_1}\) m = \(\frac{y - 3}{x}\) \(\geq\) \(\frac{-3}{2}\) 2(y - 3) \(\geq\) - 3x = 2y - 6 \(\geq\) - 3x = 2y + 3x \(\geq\) 6 ; x \(\geq\) 0, y \(\geq\) 0

o) = 90o (angle substended by a semi-circle) ∴ o o - 30o = 60o o - 35o = 55o o + 55o) = 180 - 115o = 65o o(angles in the same segment) o o + 35o = 180o o o - 155o = 25o

A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20 = \(\frac{1}{2}\) x 8 x h = 4h h = \(\frac{20}{4}\) = 5cm A\(\bigtriangleup\)(PQTS) = L x H A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS 20 + 50 = 70cm\(^2\) ALTERNATIVE METHOD A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28 = 70cm\(^2\)

< TXR = < QTR = 40o (alternate segment) < TWR = < TXR = 40o(Angles in the same segments) < YXR = 40o + 32o = 72o < YXR + < YTR = 180o(Supplementary) 72o + < YTR = 180o < YTR = 180o - 72o = 108o

\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\) \(\frac{15}{20} = \sin x\) sin x = \(\frac{15}{20} = \frac{3}{4}\) N.B x = < PTR

Length of chord = \(2r \times \sin(\frac{\theta}{2})\) = \( 2 \times \sqrt{3} \times \sin(\frac{60}{2})\) = \(2 \times \sqrt{3} \times \frac{1}{2}\) = \(\sqrt{3}\) cm.

Let the monthly expenditure on school fees be x let the monthly expenditure on housing be 2x; angle of housing and school fee in a pie chart = 360o - (120o + 90o) = 360o - 210o = 150o Angle of housing in a pie-chart = \(\frac{2}{3}\) x 150 = 100 \(\frac{100}{360}\) x 7200 = N2,000

\(\begin{array}{c|c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline f & 1 & 9 & 4 & 7 & 10 & 8 & 7 & 9 & 8 & 2 & 1\end{array}\) no pupils who failed the test = 1 + 3 + 4 + 7 + 10 = 25 5 of pupils who fail = \(\frac{25}{60}\) x 100% = 41.70%