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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%

(a) (b) A = {0, 1, 2, 3, 4, 5, 6, 7} B = {4, 6, 8, 10, 12} C = {2, 3, 5, 7} U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (i) \(B \cup C = {2, 3, 4, 5, 6, 7, 8, 10, 12}\) \((B \cup C)' = {0, 1, 9, 11}\) (ii) \(A' = {8, 9, 10, 11, 12}\) \(A' \cap B \cap C = { }\)

(a)(i) \(\log_{10} 35 = \log_{10} (7 \times 5) = \log_{10} 7 + \log_{10} 5\) = \(0.8451 + 0.6990 = 1.5441\) (ii) \(\log_{10} 2.8 = \log_{10} (\frac{4 \times 7}{10} = \log_{10} 4 + \log_{10} 7 - \log_{10} 10\) = \(\log_{10} 2^{2} + \log_{10} 7 - 1\) = \(2\log_{10} 2 + \log_{10} 7 - 1 = 2(0.3010) + 0.8451 - 1\) = \(0.4471\) (b) \(N^{0.8942} = 2.8\) \(0.8942 \log N = \log 2.8\) \(\log N = \frac{\log 2.8}{0.8942} = \frac{0.4471}{0.8942}\) \(\log N = 0.5\)

(a) \(2Ax - Kx^{2}\) When x = 1, \(2A(1) - K(1^{2}) = 7\) \(2A - K = 7 ... (1)\) When x = 2, \(2A(2) - K(2^{2}) = 4\) \(4A - 4K = 4 \implies A - K = 1 ... (2)\) From (2), \(A = 1 + K\) (1) becomes : \(2(1 + K) - K = 7\) \(2 + 2K - K = 7 \implies 2 + K = 7\) \(\therefore K = 5\) \(A = 1 + K = 1 + 5 = 6\) \(A, K = 6, 5\) (b) \(x^{2} - 3x - 1 = 0\) \(a = 1, b = -3, c = -1\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\) \(x = \frac{3 \pm \sqrt{9 + 12}}{2}\) \(x = \frac{3 \pm \sqrt{21}}{2} = \frac{3 \pm 4.583}{2}\) \(x = \frac{3 + 4.583}{2} ; x = \frac{3 - 4.583}{2}\) \(x = \frac{7.583}{2} = 3.7915 \approxeq 3.8\) (to 1 d.p) \(x = \frac{-1.583}{2} = -0.7915 \approxeq -0.8\) (to 1 d.p)

Let one of the sides be x metres. The other is (x + 1.5) metres. (a) \(Area = Length \times breadth\) \(x(x + 1.5) = 13.5\) \(x^{2} + 1.5x = 13.5 \implies x^{2} + 1.5x - 13.5 = 0\) \(x^{2} + 4.5x - 3x - 13.5 = 0 \implies x(x + 4.5) - 3(x + 4.5) = 0\) \((x - 3)(x + 4.5) = 0 \implies x = 3 ; x = -4.5\) Since x cannot be negative, therefore x = 3m. The longer side = (3 + 1.5) = 4.5m (b) \(N250 = 1m^{2}\) \(N2000 = \frac{2000}{250} = 8m^{2}\) \(\therefore \text{The area of the floor that will be covered by the carpet} = 8m^{2}\)