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a = 50o (Alt. ∠s) Bearing of Q from P = 180o + a = 180o + 50o = 230o

Tan60o = h/55 h = 55Tan60o h 55 x √3 ; h = 55√3m

Mean = ∑fx/∑f = 35/20 = 13/4 = 1.75

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.

0 - 4; Mid Values 0 + 4/2 = 2 5 - 9; 5 + 9/2 = 7 10 - 14; 10 + 14/2 = 12 15 - 19; 15 + 19/2 = 17 The class interval is 5 class boundaries are 0.5, 4.5, 9.5, 14.5, 19.5

Q1 represents the value below which 25% of the data falls, while Q3 indicates the value below which 75% of the data falls (or above which 25% falls). Consequently, the percentage of data outside the interval [Q1, Q3] is 50%, composed of 25% on each side. This principle applies to any distribution, as quartiles are determined by order statistics and cumulative probabilities, independent of the distribution's shape

Total balls = 12 RG or GR \(\frac{3}{12}\) x \(\frac{5}{11}\) + \(\frac{5}{12}\) x \(\frac{3}{11}\) = \(\frac{5}{44}\) + \(\frac{5}{44}\) =\(\frac{10}{44}\) =\(\frac{5}{22}\)

18 + 14 + 20 + 16/100 = 17/25

(a) 20 kg at N5.00 each = N100.00 10 kg at N3.00 each = N30.00 5 kg at N2.00 each = N10.00 5 kg at 50k each = N2.50 Total = N 142.50. (b) Selling price = N142.50 Cost price = N120.00 \(\therefore\) Profit = N(142.50 - 120.00) = N22.50. (c) \(% profit = \frac{22.5}{120} \times 100%\) = \(\frac{45}{240} \times 100% = 18.75%\)

(a) \(6y + 5x = -12 ... (1)\) \(4y - 3x = 11 ... (2)\) Multiply (1) by 2 and (2) by 3, \(12y + 10x = -24 ... (1a)\) \(12y - 9x = 33 ... (2a)\) \((1a) - (2a) : 19x = -57 \implies x = -3\) \(4y - 3(-3) = 11 \implies 4y + 9 = 11\) \(4y = 2 \implies y = \frac{1}{2}\) (b) \(/BC/^{2} = 36^{2} + 48^{2}\) \(/BC/^{2} = 1296 + 2304 = 3600\) \(/BC/ = \sqrt{3600} = 60 cm\) \(/AC/^{2} = /BC/^{2} + /AB/^{2}\) \((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\) \(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\) \(96y = 3600 + 1296 - 2304 = 2592\) \(96y = 2592 \implies y = \frac{2592}{96} = 27 cm\)

(a) Marks Class boundary Freq Cum freq 0 - 9 0 - 9.5 4 4 10 - 19 9.5 - 19.5 5 9 20 - 29 19.5 - 29.5 6 15 30 - 39 29.5 - 39.5 12 27 40 - 49 39.5 - 49.5 8 35 50 - 59 49.5 - 59.5 5 40 (b) Mode = 35.5 (c) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{p}) \times c}{f_{m}}\) where \(L_{1}\) = lower class boundary of median class = 29.5 \(N = \sum f = 40\) ; \(\sum f_{p}\) = cumulative frequency before median class = 15. \(f_{m}\) = frequency of median class = 12, c = class interval = 10. Median = \(29.5 + \frac{(\frac{40}{2} - 15) \times 10}{12}\) = \(29.5 + \frac{(20 - 15) \times 10}{12}\) = \(29.5 + 4.17 = 33.67\)

(a) \(\log_{10} 0.72 =\log_{10} (\frac{72}{100})\) = \(\log_{10} 72 - \log_{10} 100\) = \(72 = 2^{3} \times 3^{2}\) \(\log_{10} 72 = \log_{10} 2^{3} + \log_{10} 3^{2} = 3\log_{10} 2 + 2 \log_{10} 3\) = \(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\) \(\therefore \log_{10} 0.72 \equiv \log_{10} 72 - \log_{10} 100\) = \(1.8572 - 2 = -0.1428\) (b) \(\tan 39^\circ = \frac{20}{AB} \) = \(\frac{20}{\tan 39^\circ}\) \(AB = 24.698 m \approxeq 25m\)

(a) \(< BAD = 106^\circ\) (Given) \(/AD/ = /AB/\) (given) \(\therefore < ADC = \frac{180^\circ - 106^\circ}{2} = 37^\circ\) (base angles of an isosceles triangle) (b) Total card = 20 (i) P(card picked is E) = \(\frac{1}{20}\) (ii) P(B and A or C or D or E) or P(A or C or D or E and B) = \(\frac{4}{20} \times \frac{3 + 7 + 5 + 1}{19} + \frac{3 + 7 + 5 + 1}{20} \times \frac{4}{19}\) = \(\frac{64}{380} + \frac{64}{380} = \frac{32}{95}\)

(a) \(\frac{0.45 \times 0.91}{0.0117} = \frac{45 \times 91}{117}\) = \(5 \times 7 = 35\) (b) Halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\) = \(\frac{1\frac{6}{7} + 2\frac{11}{28}}{2}\) = \((\frac{13}{7} + \frac{67}{28}) \times \frac{1}{2}\) = \(\frac{119}{28} \times \frac{1}{2} \) = \(\frac{119}{56} = 2\frac{7}{56} = 2\frac{1}{8}\) (c) Let the interior angle = 5x^\circ and the exterior angle = x^\circ. \(\therefore 5x^\circ + x^\circ = 180^\circ\) \(6x^\circ = 180^\circ \implies x = 30^\circ\) Exterior angle = \(\frac{360^\circ}{n} \implies 30^\circ = \frac{360^\circ}{n}\) \(n = \frac{360^\circ}{30^\circ} = 12\) The polygon has 12 sides. (d) Length of pipe = 20 cm Internal diameter = 6 cm ; radius = 3cm External diameter = 8 cm ; radius = 4cm Let the external radius = R and internal radius = r \(\therefore\) Volume of material used = \(\pi R^{2} l - \pi r^{2} l\) Where l = length. \(Volume = \pi l (R^{2} - r^{2}) cm^{3}\) = \(\frac{22}{7} \times 20 (4^{2} - 3^{2}) = \frac{22}{7} \times 20 \times 7 = 440 cm^{3}\)