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(a) (b) \(x + 45 + x + 11 = 120 \implies 2x + 56 = 120\) \(2x = 120 - 56 = 64\) \(x = 32\) Therefore, the number of customers who bought bags = 32. (b) the number of customers that bought shoes = 11 + 32 + 45 = 88. (c) \(P(\text{a random customer bought bags}) = \frac{\text{no of customers that bought bags}}{\text{total no of customers}}\) No of customers that bought bags = 32 + 45 = 77 \(P(\text{customer bought bag}) = \frac{77}{120}\).

(a) Worth of raw materials = N30,450.00 Worth of raw material to get processed materials = N87.00 Amount of raw material to get processed material = \(\frac{30,450}{87}\) Time required for direct labour = 3 hours Amount that will be budgeted for direct labour at N18.25 per hour = \(3 \times 18.25 \times \frac{30,450}{87}\) = \(\frac{1667137.5}{87}\) = N19,162.50 (b) Total amount invested \(N(x + y) = N8,000,000 ....(1)\) Interest from bank M = \(\frac{x \times 4 \times 6}{100} = 0.24x\) Interest from bank N = \(\frac{y \times 4 \times 8}{100} = 0.32y\) \(0.24x + 0.32y = 2,320,000 .... (2)\) (i) Multiplying (1) by 0.24 in order to eliminate x, we have \(0.24(x + y) = 0.24(8,000,000.00)\) \(0.24x + 0.24y = 1,920,000.00 .... (3)\) (2) - (3) : \(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\) \(0.08y = 400,000 \implies y = \frac{400,000}{0.08} = N5,000,000\) \(x = 8,000,000 - y\) = \(8,000,000 - 5,000,000\) = \(N3,000,000\) Therefore, x = N3,000,000 and y = N5,000,000. (ii) Interest paid by bank N = 0.32y = \(0.32 \times N5,000,000\) = \(N1,600,000\)

(a) x -3 -2 -1 0 1 2 3 4 5 6 y 30 13 0 -9 -14 -15 -12 -5 6 21 (b) (c) (i) From the graph, the roots of the equation 2x\(^2\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\(^2\) - 7x - 9. The roots of the equation 2x\(^2\) - 7x = 26 : -2.2 and 5.8. (ii) The coordinates of the minimum point of y : 1.8, -15. (iii) The range of values for which 2x\(^{2}\) - 7x < 9 : - 0.9 < x < 4.5.

Mark (x) 1 2 3 4 5 Total Frequency (f) m + 2 m - 1 2m - 3 m + 5 3m - 4 8m - 1 fx m + 2 2m - 2 6m - 9 4m + 20 15m - 20 28m - 9 (a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) \(\frac{75}{23} = \frac{28m - 9}{8m - 1}\) \(75(8m - 1) = 23(28m - 9) \implies 600m - 75 = 644m - 207\) \(-75 + 207 = 644m - 600m\) \(132 = 44m \implies m = 3\) (b)(i) Interquartile range = Third quartile - First quartile Frequency = 8(3) - 1 = 24 - 1 = 23 \(Q_{3} = \frac{3}{4} \times 23 = 17.25th\) position = 4 \(Q_{1} = \frac{1}{4} \times 23 = 5.75th\) position = 2 Interquartile range : 4 - 2 = 2 (ii) P(at least 4 marks) = \(\frac{(m + 5 + 3m - 4)}{23} = \frac{4m + 1}{23}\) = \(\frac{4(3) + 1}{23} \) = \(\frac{13}{23}\)

(ii) (a) < RTS = 65^\circ - 34^\circ (exterior angles of triangle PST) = 31^\circ (ii) (b) < SRP = 65^\circ (angles in alternate segment are equal) \(\therefore < SRP = 180^\circ - 65^\circ = 115^\circ\). (b) If < XYZ = 128^\circ then < XZY = 128^\circ + 20^\circ = 148^\circ < XZV = 180^\circ - 148^\circ = 32^\circ < YVZ = < VYZ = 52^\circ (Isoceles triangle) < WYZ = 52^\circ - 32^\circ = 20^\circ

(a) \(\sin x = \frac{5}{13}\) Using SOHCAHTOA, then Opp = 5 and Hyp = 13 \(13^{2} = 5^{2} + Adj^{2}\) \(169 = 25 + Adj^{2}\) \(Adj = \sqrt{169 - 25} = \sqrt{144} = 12\) \(\cos x = \frac{12}{13}\) \(\tan x = \frac{5}{12}\) \(\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - 2(\frac{5}{13})}{2(\frac{5}{12})}\) = \(\frac{\frac{2}{13}}{\frac{5}{6}}\) = \(\frac{12}{65}\) (b) Taking \(\Delta LPB\), Using Pythagoras theorem, \(/BP/^{2} = /BL/^{2} - /LP/^{2}\) \(i.e. /BP/ = \sqrt{12^{2} - 9.6^{2}}\) = \(\sqrt{144 - 92.16} = \sqrt{51.84} = 7.2m\) \(/PA/ = 10m - 7.2m = 2.8m\) \(/LA/^{2} = x^{2}\) \(x^{2} = /LP/^{2} + /PA/^{2}\) \(x^{2} = 9.6^{2} + 2.8^{2}\) = \(92.16 + 7.84\) \(x^{2} = 100 \implies x = 10m\) (ii) \(\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286\) \(\theta_{y} = \tan^{-1} (3.4286) \approxeq 74^\circ\)

(a) \(Time = \frac{2}{3} hour = \frac{2}{3} \times 60 = 40 mins\) 4 students will take \(\frac{8 \times 40}{4} = 80 minutes\) \(Time = \frac{1}{3} hour = \frac{1}{3} \times 60 = 20 minutes\) It will take 20 minutes to fill \(\frac{12 \times 20}{80} = 3 tanks\) (b) \(\tan (\frac{\theta}{2}) = \frac{10}{12}\) \(\frac{\theta}{2} = \tan^{-1} (0.8337) = 39.8^\circ\) \(\theta = 39.8^\circ \times 2 = 79.6^\circ\) (ii) Perimeter of minor segment = Length of chord + length of arc But \(r^{2} = 12^{2} + 10^{2}\) = \(144 + 100 = 244\) \(r = \sqrt{244} = 15.62 cm\) Length of arc = \(\frac{79.6}{360} \times 2 \times 3.142 \times 15.62 = 21.71cm\) Perimeter of minor segment = 20 + 21.71 = 41.71cm

(a) \(3y^{2} - 5y + 2 = 0\) \(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\) \(y^{2} - \frac{5}{3}y = \frac{-2}{3}\) \(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\) \(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\) \((y - (\frac{5}{6}))^{2} = \frac{1}{36}\) Taking square root of both sides, \(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\) \(y - \frac{5}{6} =\pm {\frac{1}{6}}\) \(y = \frac{5}{6} \pm \frac{1}{6}\) \(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\). (b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) \(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\) \(m + 2n = 2 ... (1)\) \(4m + 2n = 3 ... (2)\) Multiplying (1) by 4, we have \(4m + 8n = 8 ... (3)\) (3) - (1) : \(8n - 2n = 8 - 2 = 6\) \(6n = 6 \implies n = 1\) \(m + 2n = m + 2(1) = m + 2 = 2\) \(m = 2 - 2 = 0\) \(x + 2y = 1 ... (1)\) \(4x + y = 4 ... (2)\) Multiply (1) by 4, \(4x + 8y = 4 ... (3)\) (3) - (2) : \(8y - y = 4 - 4 = 0\) \(\implies y = 0\) \(x + 2y = x + 2(0) = x + 0 = 1\) \(x = 1 - 0 = 1\) \(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

(a) (i) \(3 * \frac{2}{5} = \frac{3 + \frac{2}{5}}{2}\) = \(\frac{\frac{17}{5}}{2}\) = \(\frac{17}{10}\) (ii) \(8 * y = 8\frac{1}{4}\) \(\frac{8 + y}{2} = \frac{33}{4}\) \(66 = 32 + 4y \implies 4y = 66 - 32 = 34\) \(y = \frac{34}{4} = 8.5\) (b) \(\overrightarrow{AP} = \frac{1}{2}(\overrightarrow{AB})\) = \(\frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) = \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) Midpoint = \(\overrightarrow{AP} + \overrightarrow{PC} = \overrightarrow{AC}\) \(\overrightarrow{PC} = \overrightarrow{AC} - \overrightarrow{AP}\) = \(\begin{pmatrix} 3 \\ -8 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\) = \(\begin{pmatrix} 5 \\ -11 \end{pmatrix}\) \(\overrightarrow{CP} = -1 \times \begin{pmatrix} 5 \\ -11 \end{pmatrix}\) = \(\begin{pmatrix} -5 \\ 11 \end{pmatrix}\)

Look through the equations of motion

Electromagnetic waves do not require any material medium, but sound waves is a mechanical wave that require a material medium

Electric Field Intensity (E) = \(\frac{v}{d}\) = \(\frac{\text{potential difference}}{\text{distance}}\) = \(\frac{6.5v}{35cm}\) = \(\frac{6.5}{35 \times 10^{- 2}}\) = 18.57NC\(^{-1}\)

Soldiers march disorderly while crossing a bridge to prevent resonance because when soldiers march in orderly manner across a bridge, they generate a rhythmic oscillation of wave on the bridge and at a certain point, would start oscillation to the same rhythm as that of the marching steps. This oscillation would reach a maximum peak when the bridge can no longer sustain its own strength and hence collapses. So for this reasons, soldier are ordered to march disorderly while crossing a bridge.

Upthrust = change in weight density = \(\frac{\text{mass}}{\text{volume}}\) Mass of liquid displaced = Density of Liquid \times Volume; = \(10^{3} \times 50 \times 10^{−6}m^{3}\) = 0.05kg (Note that the volume of \(50cm^{3}\) was converted to \(m^{3}\) by multiplying by \(10^{6}\)) i.e \(50cm^{3} = 50 \times 10^{−6}m^{3}\) Since mass displaced = 0.05kg Upthrust = mg = 0.05 \times 10 = 0.5N

Cathode rays are produced in a discharge tube under Low pressure and high voltage