WAEC SSCE - Further Mathematics - 2009 (Essay)

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\(A(-2,1),\ B(4,-2),\ C(1,8)\). \(D(x,y)\) is the foot of the perpendicular from \(A\) to \(BC\).

(a) Direction of \(BC\): \(\vec{BC} = C - B = (1 - 4,\ 8 - (-2)) = (-3,\ 10)\). Vector \(\vec{AD} = (x + 2,\ y - 1)\). Since \(AD \perp BC\), their dot product is zero:

\[(x + 2)(-3) + (y - 1)(10) = 0\] \[-3x - 6 + 10y - 10 = 0 \;\Rightarrow\; 3x - 10y + 16 = 0\]

The equation connecting \(x\) and \(y\) is \(3x - 10y + 16 = 0\).

(b) \(|\vec{BC}| = \sqrt{(-3)^2 + 10^2} = \sqrt{109}\). Unit vector in the direction of \(BC\):

\[\hat{BC} = \frac{1}{\sqrt{109}}\begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -\tfrac{3}{\sqrt{109}} \\[4pt] \tfrac{10}{\sqrt{109}} \end{pmatrix}\]

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(a) Write each combination out:

\[{}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6},\qquad {}^{n-1}C_3 = \frac{(n-1)(n-2)(n-3)}{6}\] \[{}^{n+1}C_3 - {}^{n-1}C_3 = \frac{n-1}{6}\Big[(n+1)n - (n-2)(n-3)\Big]\] \[= \frac{n-1}{6}\Big[(n^2 + n) - (n^2 - 5n + 6)\Big] = \frac{n-1}{6}(6n - 6) = (n - 1)^2\]

So the expression simplifies to \((n - 1)^2\).

(b) A six has \(p = \dfrac16,\ q = \dfrac56\), with \(n = 5\) throws (binomial).

(i) At most two sixes \(= P(0) + P(1) + P(2)\):

\[P(0) = \left(\tfrac56\right)^5 = \tfrac{3125}{7776},\quad P(1) = \binom{5}{1}\tfrac16\left(\tfrac56\right)^4 = \tfrac{3125}{7776},\quad P(2) = \binom{5}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^3 = \tfrac{1250}{7776}\] \[P(\le 2) = \frac{3125 + 3125 + 1250}{7776} = \frac{7500}{7776} \approx 0.965\]

(ii) Exactly three sixes:

\[P(3) = \binom{5}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^2 = 10\cdot\frac{1}{216}\cdot\frac{25}{36} = \frac{250}{7776} \approx 0.032\]

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Reading the diagram, the 6N force acts due East (direction 090°). Measuring anticlockwise from this 6N line, the other forces sit at these angles:

• 6N along East: \(0^\circ\)
• 8N: \(55^\circ\) above the 6N line, so \(55^\circ\)
• 4N (upper-left): a further \(100^\circ\) beyond the 8N, so \(55^\circ+100^\circ=155^\circ\)
• 4N (lower-right): \(25^\circ\) below East, so \(-25^\circ\) i.e. \(335^\circ\)
• 5N (lower-left): \(130^\circ\) round from the lower 4N, i.e. \(-25^\circ-130^\circ=-155^\circ\) i.e. \(205^\circ\)

(a) Resolve each force into components along the 6N direction (\(x\)) and perpendicular to it (\(y\)).

\[\sum F_x = 6\cos0^\circ+8\cos55^\circ+4\cos155^\circ+5\cos205^\circ+4\cos335^\circ\]

\[\sum F_x = 6+4.589-3.625-4.532+3.625 = 6.057\ \text{N}\]

\[\sum F_y = 6\sin0^\circ+8\sin55^\circ+4\sin155^\circ+5\sin205^\circ+4\sin335^\circ\]

\[\sum F_y = 0+6.553+1.690-2.113-1.690 = 4.440\ \text{N}\]

Magnitude of the resultant:

\[R=\sqrt{(\sum F_x)^2+(\sum F_y)^2}=\sqrt{6.057^2+4.440^2}=\sqrt{36.69+19.71}=\sqrt{56.40}\]

\[R\approx 7.51\ \text{N}\]

(b) Direction of the resultant. Both components are positive, so the resultant lies above the 6N (East) line.

\[\theta=\tan^{-1}\!\left(\frac{\sum F_y}{\sum F_x}\right)=\tan^{-1}\!\left(\frac{4.440}{6.057}\right)=\tan^{-1}(0.7332)\approx 36.2^\circ\]

The resultant is about 7.51 N acting at \(36.2^\circ\) above the 6N direction. Since the 6N points along a bearing of \(090^\circ\), this direction corresponds to a bearing of \(090^\circ-36.2^\circ\approx \mathbf{054^\circ}\).

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From \(\log_2 x - \log_2 y = 2\):

\[\log_2\frac{x}{y} = 2 \;\Rightarrow\; \frac{x}{y} = 2^2 = 4 \;\Rightarrow\; x = 4y\]

From \(\log_2(x - 2y) = 3\):

\[x - 2y = 2^3 = 8\]

Substitute \(x = 4y\):

\[4y - 2y = 8 \;\Rightarrow\; 2y = 8 \;\Rightarrow\; y = 4,\quad x = 16\]

\(x = 16,\ y = 4\). Check: \(\log_2 16 - \log_2 4 = 4 - 2 = 2\) and \(\log_2(16 - 8) = \log_2 8 = 3\).

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(a)(i) With \( A(1,5),\ B(3,9),\ C(-1,1) \):

\( \vec{AB} = B-A = (2,\ 4) \) and \( \vec{BC} = C-B = (-4,\ -8) \).

Since \( \vec{BC} = -2\,\vec{AB} \), the two vectors are parallel and share the common point B. Therefore A, B and C are collinear.

(ii) \( |AB| = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt5 \) and \( |BC| = \sqrt{(-4)^2+(-8)^2} = \sqrt{80} = 4\sqrt5 \).

\[ |AB| : |BC| = 2\sqrt5 : 4\sqrt5 = 1 : 2. \]

(b) Let the beam XY, length 24 m, lie horizontally. String tensions: \(T_X\) at X and \(T_M\) at M, where M is 4 m from Y, i.e. 20 m from X. The beam weight \(10g\) acts at the centre, 12 m from X; the 50 kg load acts at N, 8 m from X (take \(g = 10\text{ ms}^{-2}\)).

Vertical equilibrium: \( T_X + T_M = (10+50)g = 60g \).

Moments about X: \[ T_M(20) = 10g(12) + 50g(8) = 120g + 400g = 520g \Rightarrow T_M = 26g = 260\text{ N}. \]

Then \( T_X = 60g - 26g = 34g = 340\text{ N} \).

The tension at X is \(340\text{ N}\) and the tension at M is \(260\text{ N}\).

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Expand and gather to standard form:

\[(x+1)(x+2) = k(3x+7)\] \[x^2 + 3x + 2 = 3kx + 7k\] \[x^2 + (3 - 3k)x + (2 - 7k) = 0\]

For equal roots the discriminant is zero:

\[(3 - 3k)^2 - 4(2 - 7k) = 0\] \[9 - 18k + 9k^2 - 8 + 28k = 0\] \[9k^2 + 10k + 1 = 0\] \[(9k + 1)(k + 1) = 0\]

Hence \(k = -\dfrac{1}{9}\) or \(k = -1\).

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(a) Five ordinates means four strips of width \(h = \dfrac{1 - 0}{4} = 0.25\). With \(f(x) = \dfrac{3}{1 + x^2}\):

\(x\)	0	0.25	0.5	0.75	1
\(f(x)\)	3.0000	2.8235	2.4000	1.9200	1.5000

\[\int_0^1 f\,dx \approx \frac{h}{2}\Big[f_0 + f_4 + 2(f_1 + f_2 + f_3)\Big] = \frac{0.25}{2}\Big[3 + 1.5 + 2(2.8235 + 2.4 + 1.92)\Big]\] \[= 0.125\big[4.5 + 2(7.1435)\big] = 0.125(18.787) = 2.348\ (\text{4 s.f.})\]

(b) \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\). First \(A^2\):

\[A^2 = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix}\] \[A^2 + A + 2I = \begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}\]

Image of \((1, 2)\):

\[\begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}\]

The image is \((21,\ 37)\).

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(a) For a G.P., \(T_3 = ar^2 = 2\) and \(T_6 = ar^5 = 54\).

(i) Dividing: \(\dfrac{ar^5}{ar^2} = \dfrac{54}{2} \Rightarrow r^3 = 27 \Rightarrow r = 3\).

(ii) \(ar^2 = 2 \Rightarrow 9a = 2 \Rightarrow a = \dfrac{2}{9}\).

(iii) Sum of first 10 terms:

\[S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{\tfrac{2}{9}(3^{10} - 1)}{3 - 1} = \frac{\tfrac{2}{9}(59049 - 1)}{2} = \frac{59048}{9} \approx 6561\]

(b) In \((1 + 2x)^n\): coefficient of \(x^3 = \binom{n}{3}2^3\); coefficient of \(x^4 = \binom{n}{4}2^4\).

\[\frac{\text{coeff }x^4}{\text{coeff }x^3} = \frac{16\binom{n}{4}}{8\binom{n}{3}} = 2\cdot\frac{n-3}{4} = \frac{n-3}{2}\]

Setting this equal to \(\dfrac{3}{1}\):

\[\frac{n - 3}{2} = 3 \;\Rightarrow\; n - 3 = 6 \;\Rightarrow\; n = 9\]

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\(\mathbf{r} = 4t\,\mathbf{i} + (12 - 3t)\,\mathbf{j}\).

(a) Velocity is the time derivative:

\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = 4\,\mathbf{i} - 3\,\mathbf{j}\ (\text{ms}^{-1})\]

(b) Positions at the two times:

\[\mathbf{r}(0) = 0\,\mathbf{i} + 12\,\mathbf{j},\qquad \mathbf{r}(5) = 20\,\mathbf{i} + (12 - 15)\,\mathbf{j} = 20\,\mathbf{i} - 3\,\mathbf{j}\]

Displacement \(= \mathbf{r}(5) - \mathbf{r}(0) = 20\,\mathbf{i} - 15\,\mathbf{j}\).

\[|\text{displacement}| = \sqrt{20^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25\ \text{units}\]

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(a) Sketching the two curves on the same axes.

Curve 1: \(y = 6 - x - x^{2}\)

This is an inverted (\(\cap\)) parabola since the coefficient of \(x^{2}\) is negative.

• y-intercept: when \(x = 0,\ y = 6\), so it passes through \((0,\,6)\).
• x-intercepts: when \(y = 0,\ 6 - x - x^{2} = 0 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x+3)(x-2) = 0\), giving \(x = -3\) and \(x = 2\), i.e. the points \((-3,\,0)\) and \((2,\,0)\).
• Turning point: \(\dfrac{dy}{dx} = -1 - 2x = 0 \Rightarrow x = -\tfrac12\). Then \(y = 6 - (-\tfrac12) - (-\tfrac12)^{2} = 6\tfrac14\). Maximum at \(\left(-\tfrac12,\ 6\tfrac14\right)\).

Curve 2: \(y = 3x^{2} - 2x + 3\)

This is an upright (\(\cup\)) parabola since the coefficient of \(x^{2}\) is positive.

• y-intercept: when \(x = 0,\ y = 3\), so it passes through \((0,\,3)\).
• x-intercepts: \(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 = -32 < 0\), so the curve does not cut the x-axis.
• Turning point: \(\dfrac{dy}{dx} = 6x - 2 = 0 \Rightarrow x = \tfrac13\). Then \(y = 3(\tfrac13)^{2} - 2(\tfrac13) + 3 = 2\tfrac23\). Minimum at \(\left(\tfrac13,\ 2\tfrac23\right)\).

Plotting both on the same axes, the \(\cap\) parabola sits above the \(\cup\) parabola between their two crossing points, enclosing the finite shaded region:

(b) x-coordinates of the points of intersection.

At the points of intersection the two y-values are equal:

\[6 - x - x^{2} = 3x^{2} - 2x + 3\]\[0 = 3x^{2} - 2x + 3 - 6 + x + x^{2}\]\[0 = 4x^{2} - x - 3\]

Factorising: \(4x^{2} - 4x + 3x - 3 = 4x(x-1) + 3(x-1) = (4x + 3)(x - 1) = 0\).

\[x = -\frac{3}{4} \quad\text{or}\quad x = 1\]

So the curves intersect where \(x = -\tfrac34\) and \(x = 1\).

(c) Area of the finite region bounded by the two curves.

Between \(x = -\tfrac34\) and \(x = 1\), the curve \(y = 6 - x - x^{2}\) lies above \(y = 3x^{2} - 2x + 3\), so the enclosed area is

\[A = \int_{-3/4}^{1}\big[(6 - x - x^{2}) - (3x^{2} - 2x + 3)\big]\,dx = \int_{-3/4}^{1}\big(3 + x - 4x^{2}\big)\,dx\]\[A = \left[\,3x + \frac{x^{2}}{2} - \frac{4x^{3}}{3}\,\right]_{-3/4}^{1}\]

At \(x = 1:\quad 3(1) + \dfrac{1}{2} - \dfrac{4}{3} = \dfrac{18 + 3 - 8}{6} = \dfrac{13}{6}\).

At \(x = -\tfrac34:\quad 3\left(-\tfrac34\right) + \dfrac{(-3/4)^{2}}{2} - \dfrac{4}{3}\left(-\tfrac34\right)^{3} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{4}{3}\cdot\dfrac{27}{64} = -\dfrac{9}{4} + \dfrac{9}{32} + \dfrac{9}{16} = -\dfrac{45}{32}\).

\[A = \frac{13}{6} - \left(-\frac{45}{32}\right) = \frac{208}{96} + \frac{135}{96} = \frac{343}{96} \approx 3.57\ \text{square units}\]

The area of the finite region bounded by the two curves is \(\dfrac{343}{96} \approx 3.57\) square units.

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(a) Scatter diagram. The ten pairs \((33,4),(31,6),(28,4),(25,10),(23,12),(22,10),(19,14),(17,15),(16,18),(14,22)\) are plotted with \(X\) on the horizontal axis and \(Y\) on the vertical axis. The points trend from the upper-left down to the lower-right, showing that \(Y\) falls as \(X\) rises.

(b) Means.

\[\bar{x}=\frac{33+31+28+25+23+22+19+17+16+14}{10}=\frac{228}{10}=22.8\] \[\bar{y}=\frac{4+6+4+10+12+10+14+15+18+22}{10}=\frac{115}{10}=11.5\]

(c) Line of best fit. A single straight line is drawn through the mean point \((\bar{x},\bar{y})=(22.8,\,11.5)\) (marked ★ on the graph) with roughly equal numbers of points on either side. Its least-squares gradient is

\(n\)	\(\sum X\)	\(\sum Y\)	\(\sum XY\)	\(\sum X^2\)
10	228	115	2293	5574

\[b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^{2}-\left(\sum X\right)^{2}}=\frac{10(2293)-228(115)}{10(5574)-228^{2}}=\frac{22930-26220}{55740-51984}=\frac{-3290}{3756}=-0.88\]

The intercept follows from \(\bar{y}=a+b\bar{x}\):

\[a=\bar{y}-b\bar{x}=11.5-(-0.88)(22.8)=11.5+20.06=31.5\]

so the line of best fit is

\[Y=31.5-0.88X.\]

(d)(i) Relationship. The line has a negative gradient, so there is a negative (inverse) correlation between \(X\) and \(Y\): as \(X\) increases, \(Y\) decreases.

(d)(ii) Value of \(Y\) when \(X=24\). Reading up from \(X=24\) to the line of best fit and across to the \(Y\)-axis, or substituting into the equation:

\[Y=31.5-0.88(24)=31.5-21.1=10.4\]

so \(Y\approx 10.5\) (about 10 to 11 from a hand-drawn line).

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Total selections of \(5\) from \(14\) people: \(\binom{14}{5} = 2002\).

(a) 3 girls and 2 boys:

\[\binom{8}{3}\binom{6}{2} = 56\times15 = 840\] \[P = \frac{840}{2002} = \frac{60}{143} \approx 0.420\]

(b) All girls or all boys:

\[\binom{8}{5} + \binom{6}{5} = 56 + 6 = 62\] \[P = \frac{62}{2002} = \frac{31}{1001} \approx 0.031\]

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(a) Let \(u = 5 - x^2\), so \(du = -2x\,dx\) and \(x\,dx = -\tfrac12\,du\).

\[\int \frac{x}{\sqrt{5 - x^2}}\,dx = -\frac12\int u^{-1/2}\,du = -\sqrt{u} = -\sqrt{5 - x^2}\] \[\int_{1}^{2}\frac{x}{\sqrt{5 - x^2}}\,dx = \Big[-\sqrt{5 - x^2}\Big]_1^2 = -\sqrt{1} + \sqrt{4} = -1 + 2 = 1\]

(b)(i) Expanding along the first row:

\[\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix} = 2(1\cdot(-3) - (-2)\cdot2) + 3(0\cdot(-3) - (-2)\cdot1) + 1(0\cdot2 - 1\cdot1)\] \[= 2(1) + 3(2) + 1(-1) = 2 + 6 - 1 = 7\]

(ii) The coefficient matrix of the system is exactly this determinant, so \(\Delta = 7\). By Cramer's rule (replacing each column with \((10, -7, -9)^T\)):

\[\Delta_x = \begin{vmatrix} 10 & -3 & 1 \\ -7 & 1 & -2 \\ -9 & 2 & -3 \end{vmatrix} = 14,\quad \Delta_y = \begin{vmatrix} 2 & 10 & 1 \\ 0 & -7 & -2 \\ 1 & -9 & -3 \end{vmatrix} = -7,\quad \Delta_z = \begin{vmatrix} 2 & -3 & 10 \\ 0 & 1 & -7 \\ 1 & 2 & -9 \end{vmatrix} = 21\] \[x = \frac{14}{7} = 2,\quad y = \frac{-7}{7} = -1,\quad z = \frac{21}{7} = 3\]

\(x = 2,\ y = -1,\ z = 3\). Check: \(2(2) - 3(-1) + 3 = 10\) and \((-1) - 2(3) = -7\).

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(a) K divides LM with LK:KM = 2:3, so the position vector of K is the weighted mean \[ \vec{K} = \frac{3\vec{L} + 2\vec{M}}{5}. \] Thus \[ \vec{K} = \frac{3(5\mathbf{i}+6\mathbf{j}) + 2(13\mathbf{i}+4\mathbf{j})}{5} = \frac{(15+26)\mathbf{i} + (18+8)\mathbf{j}}{5} = \frac{41\mathbf{i}+26\mathbf{j}}{5}. \] Hence \( \vec{K} = 8.2\mathbf{i} + 5.2\mathbf{j} \).

(b) Resolve each leg into east (x) and north (y) components using bearing \( \theta \): east \(= d\sin\theta\), north \(= d\cos\theta\).

\( \vec{AB} = (8\sin 60^\circ,\ 8\cos 60^\circ) = (6.928,\ 4.000) \).
\( \vec{BC} = (12\sin 130^\circ,\ 12\cos 130^\circ) = (9.193,\ -7.713) \).

\( \vec{AC} = \vec{AB}+\vec{BC} = (16.121,\ -3.713) \).

(i) \( |AC| = \sqrt{16.121^2 + (-3.713)^2} = \sqrt{259.9+13.79} = \sqrt{273.7} = 16.5\text{ km} \) (3 s.f.).

(ii) The vector points east and south, so the bearing lies between \(90^\circ\) and \(180^\circ\): \[ \text{bearing} = 90^\circ + \tan^{-1}\!\left(\frac{3.713}{16.121}\right) = 90^\circ + 12.97^\circ \approx 103^\circ. \] The bearing of C from A is \(103^\circ\) (nearest degree).

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(a) Factorise \(2x^2 + x - 6 = (2x - 3)(x + 2)\). Roots at \(x = \dfrac{3}{2}\) and \(x = -2\). Since the parabola opens upward, it is negative between the roots:

\[-2 < x < \frac{3}{2}\]

(b) Multiply numerator and denominator by the conjugate \((3\sqrt5 - \sqrt2)\):

\[\text{Denominator} = (3\sqrt5)^2 - (\sqrt2)^2 = 45 - 2 = 43\] \[\text{Numerator} = (5 - 2\sqrt{10})(3\sqrt5 - \sqrt2) = 15\sqrt5 - 5\sqrt2 - 6\sqrt{50} + 2\sqrt{20}\]

Since \(\sqrt{50} = 5\sqrt2\) and \(\sqrt{20} = 2\sqrt5\):

\[= 15\sqrt5 - 5\sqrt2 - 30\sqrt2 + 4\sqrt5 = 19\sqrt5 - 35\sqrt2\]

Therefore

\[\frac{5 - 2\sqrt{10}}{3\sqrt5 + \sqrt2} = \frac{19\sqrt5 - 35\sqrt2}{43} = -\frac{35}{43}\sqrt2 + \frac{19}{43}\sqrt5\]

So \(m = -\dfrac{35}{43}\) and \(n = \dfrac{19}{43}\).

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Let \(A = 15^{o}\) so that \(2A = 30^{o}\) and \(\tan30^{o} = \dfrac{1}{\sqrt3}\). Put \(t = \tan15^{o}\).

\[\tan30^{o} = \frac{2\tan15^{o}}{1 - \tan^2 15^{o}} \;\Rightarrow\; \frac{1}{\sqrt3} = \frac{2t}{1 - t^2}\] \[1 - t^2 = 2\sqrt3\,t \;\Rightarrow\; t^2 + 2\sqrt3\,t - 1 = 0\] \[t = \frac{-2\sqrt3 \pm \sqrt{12 + 4}}{2} = \frac{-2\sqrt3 \pm 4}{2} = -\sqrt3 \pm 2\]

Since \(15^{o}\) is acute, \(\tan15^{o} > 0\), so we take the positive value:

\[\tan15^{o} = 2 - \sqrt3\]

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First close the gaps between the stated class limits to obtain the true class boundaries, and set out the mid-values, class boundaries and the working columns for the assumed-mean method (assumed mean \(A = 43\)).

Battery life (days)	Class boundaries	Mid-value \(x\)	Frequency \(f\)	\(d = x - 43\)	\(fd\)
26 - 30	25.5 - 30.5	28	4	-15	-60
31 - 35	30.5 - 35.5	33	7	-10	-70
36 - 40	35.5 - 40.5	38	13	-5	-65
41 - 45	40.5 - 45.5	43	8	0	0
46 - 50	45.5 - 50.5	48	6	5	30
51 - 55	50.5 - 55.5	53	2	10	20
Total			40		-145

(a) Histogram

The bars are drawn contiguously on the class boundaries, with frequency on the vertical axis. Because the classes are of equal width (5 days) the bar heights equal the frequencies \(4, 7, 13, 8, 6, 2\).

(b) Mode from the histogram

The tallest bar is the modal class \(36 - 40\) (boundaries \(35.5 - 40.5\)). On the histogram two lines are drawn inside this bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. A vertical line dropped from where they cross meets the axis at the mode.

This agrees with the formula, where \(L = 35.5\), \(d_1 = 13 - 7 = 6\), \(d_2 = 13 - 8 = 5\) and \(c = 5\):

\[\text{Mode} = L + \frac{d_1}{d_1 + d_2}\,c = 35.5 + \frac{6}{6 + 5}\times 5 = 35.5 + 2.73 = 38.2\text{ days}\]

(c) Mean (assumed mean \(A = 43\))

\[\bar{x} = A + \frac{\sum fd}{\sum f} = 43 + \frac{-145}{40} = 43 - 3.625 = 39.375 \approx 39.4\text{ days}\]

(d) Percentage living less than 31 days or more than 45 days

Less than 31 days is the class \(26 - 30\): \(4\) batteries. More than 45 days covers the classes \(46 - 50\) and \(51 - 55\): \(6 + 2 = 8\) batteries.

\[\text{Number} = 4 + 6 + 2 = 12\] \[\text{Percentage} = \frac{12}{40}\times 100\% = 30\%\]

Summary: mode \(\approx 38.2\) days, mean \(\approx 39.4\) days, and \(30\%\) of the batteries last under 31 days or over 45 days.

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Let the probabilities of topping be \(M = 0.9,\ P = 0.75,\ B = 0.8\), with complements \(0.1,\ 0.25,\ 0.2\). "At least two" means exactly two or all three.

All three: \(0.9\times0.75\times0.8 = 0.54\).

Exactly two:

• \(M,P\) not \(B\): \(0.9\times0.75\times0.2 = 0.135\)
• \(M,B\) not \(P\): \(0.9\times0.25\times0.8 = 0.180\)
• \(P,B\) not \(M\): \(0.1\times0.75\times0.8 = 0.060\)

Sum of exactly two \(= 0.135 + 0.180 + 0.060 = 0.375\).

\[P(\text{at least two}) = 0.375 + 0.54 = \mathbf{0.915}\]