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Angle of sector = 108^\circ; Radius of circle = 3\(\frac{1}{2}\)cm ⇒ \(\frac{7}{2}\) cm Perimeter of sector = 108/360 \times 2\pir + 2r ⇒ \(\frac{108}{360 }\) \times 2 \times \(\frac{22}{7}\) \times \(\frac{7}{2}\) + 2(\(\frac{7}{2}\)) ⇒ 66/10 + 7 cm = 6\(\frac{3}{5}\) + 7 = 13\(\frac{3}{5}\)

∠DCB = ∠CBD = 35º (base angles of an isosceles triangle are equal) ∠ADB = 35º + 35º = 70^\circ (ext. angle of a triangle equals two opp. interior angles) ∠ABD = 90º (angle in a semicircle is a right angle) ∠BAO + ∠ADB + ∠ADB = 180º (sum of angles in a triangle) ∠BAO + 90º + 70º = 180º ∠BAO + 160º = 180º ∠BAO = 180º - 160º ∠BAO = 20º

The sides are in a common ratio ⇒ \(\frac{EC}{AE} = \frac{AB}{CD}\) \(\frac{EC}{AE} = \frac{9}{12}\) EC = \(\frac{8 \times 9}{12}\) = 6cm.

8 appears 4 times (highest frequency) → Mode = 8

Since sin x = cos (90 - x) [complementary angles] .: sin 20 = cos 70 cos 25 = sin 65 . . \(\frac{sin 20}{ cos 70} + \frac{cos 25 }{sin 65}\) =\(\frac{sin 20 }{ sin 20} + \frac{cos 25 }{cos 25}\) = 1 + 1 = 2

Total sum = 20 \times 4 = 80 Remaining sum = 80 – 23 = 57 Remaining observations = 19 Mean = \(\frac{57}{19}\) = 3

a. px - 2qx - 4qy + 2py = x(p - 2q) - 2y(2q - p) = x(p - 2q) + 2y(p - 2q) = (x + 2y)(p - 2q) b. U = {1,2,3,4,5,6,7,8,9,10} P = {1,2,4,6,10} Q = {2,3,6,9} (P ∪ Q) = {1,2,3,4,6,9,10} (P ∪ Q)\(^I\) = Elements in U but not in (P ∪ Q) (P ∪ Q)\(^I\) = {5,7,8} P\(^I\) = {3,5,7,8,9} Q\(^I\) = {1,4,5,7,8,10} P\(^I\) ∩ Q\(^I\) = {5,7,8} Evidently, (P ∪ Q)\(^I\) = P\(^I\) ∩ Q\(^I\)

a. y = c + k(\(\frac{1}{x^2}\)) where c and k are constant b. when x = 1, y = 11 ∴ 11 = c + k(\(\frac{1}{x^2}\)) = > 11 = c + k --- eqn (I) when x = 2, y = 5 ∴ 5 = c + k(\(\frac{1}{x^2}\)) = > 5 = c + k\(\frac{1}{2^2}\) --- eqn (II) Multiply through by 4) 20 = 4c + k - - - - - - - -(III) 11 = c + k --- eqn (I) 20 = 4c + k - - - - - - (III) From eqn(I) k = 11 - c Put k = 11 - c into eqn(III) 20 = 4c + k = 20 = 4c + 11 - c 20 = 3c + 11 3c = 20 - 11 = 9 3c = 9 c = 3. Put c = 3 into eqn(I) 11 = c + k 11 = 3 + k k = 11 - 3 = 8 ∴ c = 3, k = 8. When x = 4, But c = 3, K = 8 y = 3 + \(\frac{8}{4^2}\) y = 3 + \(\frac{1}{2}\) y = \(\frac{7}{2}\).

a. From the figure: |PZ| = |PQ| + |QY| + |YZ| |PQ| = |RS| = 3 cm (opposite sides of a parallelogram) |YZ| = |RS| = 3 cm (opposite sides of a parallelogram) |QY| = 2 cm (given) |PZ| = |PQ| + |QY| + |YZ| |PZ| = 3 + 2 + 3 = 8 cm b. radius of latitude r = R cos \alpha, where \alpha is the angle 56º r = 6400 cos 56^\circ = 3578.8 km < PLQ = 95º - 25^\circ = 70^\circ Length of arc PQ = \(\frac{\theta}{360}\) \times 2\pir = \(\frac{70}{360}\) \times 2 \times (\(\frac{22}{7}\)) \times 3578.8 = \(\frac{39367.2}{9}\) = 4374 km

a(i) Σfx =2 + 6 +12 + 16 + 20 + 24 + 35 + 24 +18 +10 = 167 Mean = \(\frac{Σfx}{Σf}\) =\(\frac{167}{32}\) = 5.22 ii. Median = Middle Position Value = \(\frac{\text{N}}{2}\)th position = \(\frac{32}{2}\) position = 16 th position .: Median = 5 iii. Mode = most people => highest frequency .: Mode = 7 b. People scoring at least 8 marks => frequency of x > = 8 = 3 + 1 + 2 = 6 Percentage = \(\frac{6}{32}\) x 100% = 18.75%

ai From the four-figure/ mathematical table 2sin63.35º = 2(0.8938) = 1.7876 ii. From the four-figure/ mathematical table, log cos 44.74º = log 0.7103 = \(\bar{1}\).8515 iii. log k - log(k - 2) = log 5 = log \(\frac{\text{k}}{k - 2}\) = log 5 (law of logarithms) log \(\frac{\text{k}}{k - 2}\) = log 5 \(\frac{\text{k}}{k - 2}\) = 5 cross multiply 5(k - 2) = k 5k - 10 = k 5k - k = 10 4k = 10 k = \(\frac{10}{4}\) = \(\frac{5}{2}\). b. \(\frac{3.68^2 \times 6.705}{\sqrt{0.3581}}\) = 151.8. See the steps in the diagram above.

a. p = x + ym\(^3\) ym\(^3\) = p - x m\(^3\) = \(\frac{p - x}{\text{y}}\) \(m = \sqrt[3]{\frac{p - x}{y}}\). b. x\(^2\) - 6x + 7 = 0 Using the completing the square method x\(^2\) - 6x = - 7 Add the square of half the coefficient of x to both sides x\(^2\) - 6x + (-3)\(^2\) = -7 + 9 x\(^2\) - 6x + (-3)\(^2\) = 2 (x - 3)\(^2\) = 2 x - 3 = \pm \(\sqrt{2}\) x = 3 + \(\sqrt{2}\) or 3 - \(\sqrt{2}\) x = 3 + 1.4142 or 3 - 1.4142 x = 4.1 or 1.6 to 1 decimal place. c. Let the two consecutive positive odd numbers be \(n\) and \(n + 2\). The equation for their product is: \(n(n + 2) = 195\) Expanding gives: n\(^2\) + 2n - 195 = 0 Using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 2\), and \(c = -195\): \(b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-195) = 4 + 780 = 784\) Now substituting: \(n = \frac{-2 \pm \sqrt{784}}{2} = \frac{-2 \pm 28}{2}\) This results in: \(n = \frac{26}{2} = 13 \quad (\text{positive solution})\) Thus, the two numbers are: \(13 \quad \text{and} \quad 15\)

ai. ∠Y = 270º - (195^\circ + 30^\circ) = 270^\circ - 225^\circ = 45º ∠Z = 180º - (30^\circ + 45º) = 105º Let the distance of Y from Z be k \(\frac{\text{k}}{sin30º} = \frac{20}{sin 105º}\) k sin 105º = 20 sin 30º k = \(\frac{20 sin 30^\circ}{ sin 105^\circ}\) = \(\frac{20 \times 0.5 }{sin 105^\circ}\) = \(\frac{10}{0.9659}\) = 10.35 ⇒ k = 10.4 km (3. s.f.) ii. Let the distance of Z from X be t Using the sine rule \(\frac{\text{t}}{sin 30º} = \frac{20}{sin 105º}\) t sin 105^\circ = 20 sin 45^\circ t = \(\frac{20 sin 45º}{sin 105º}\) t = 14.6km to 3. s.f bi. Latitudinal difference = 36^\circ + 36^\circ = 72^\circ Distance PQ = Length of arc PQ = \(\frac{\theta}{360}\) \times 2\piR = \(\frac{72}{360}\) \times 2 \times \(\frac{22}{7}\) \times 6400 km = \(\frac{88 \times 640}{ 7 }\) = \(\frac{56320}{7}\) = 8045.7 km = 8050 km (3 s.f.) ii. Speed of aircraft = 800 km/hr Time taken = \(\frac{\text{Total distance}}{\text{Speed}}\) = \(\frac{8050}{800}\) = 10.05 hrs

a. x f cum. frequency End point 11-20 68 68 20.5 21-30 184 252 30.5 31-40 294 546 40.5 41-50 402 948 50.5 50-60 480 1428 60.5 61-70 310 1738 70.5 71-80 164 1902 80.5 81-90 94 2000 90.5 bi. Cut-off mark = 64% ii. Pr(student scoring at least 45%) = \(\frac{1075}{2000}\) = 0.5375

a. Total ball = 5 blues + 3 blacks + 2 red balls = 10 balls Selection with replacement Pr(1st red ball) = \(\frac{2}{10}\) Pr(2nd red ball) = \(\frac{2}{10}\) Therefore, Pr(two red balls) = \(\frac{2}{10}\) x \(\frac{2}{10}\) = \(\frac{4}{100}\) = \(\frac{1}{25}\) = 0.04 b. Pr( of selecting 2 blue) = \(\frac{5}{10}\) x \(\frac{5}{10}\) = \(\frac{25}{100}\) = \(\frac{1}{4}\) Pr(of selecting 2 black balls) = \(\frac{3}{10}\) x \(\frac{3}{10}\) = \(\frac{9}{100}\) Pr( 2 blue or 2 black balls) = \(\frac{1}{4}\) + \(\frac{9}{100}\) = \(\frac{25}{100}\) + \(\frac{9}{100}\) = \(\frac{34}{100}\) = 0.34. c) 1 black, and 1 red in any other means it can be 1 black, and 1 red, and 1 black = \(\frac{3}{10}\) x \(\frac{2}{10}\) + \(\frac{2}{10}\) + \(\frac{3}{10}\) = \(\frac{6}{100}\) + \(\frac{6}{100}\) = \(\frac{12}{100}\) = 0.12.