WAEC SSCE - Further Mathematics - 2019 (Essay)

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This is a concurrent-forces (Lami's theorem) problem in which a load of 120 N is held in equilibrium by a tension \(T\) and a reaction \(R\), with the geometry giving angles of \(90^\circ\) and \(135^\circ\) opposite \(T\) and the load respectively.

(a) By Lami's theorem, \(\dfrac{T}{\sin 90^\circ}=\dfrac{120}{\sin 135^\circ}\), so

\[T=\frac{120\sin 90^\circ}{\sin 135^\circ}=\frac{120(1)}{0.7071}=169.71\ \text{N}.\]

(b) Similarly \(\dfrac{R}{\sin 135^\circ}=\dfrac{120}{\sin 135^\circ}\), giving

\[R=120\times\frac{\sin 135^\circ}{\sin 135^\circ}=120\ \text{N}.\]

Hence \(T\approx\textbf{169.71 N}\) and \(R=\textbf{120 N}\), confirming the stated working. (The full configuration/diagram was not supplied, so the angle assignments are taken from the given ratios.)

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Let \(f(x)=3x^2+2x-1\). By first principles \(f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).

\[f(x+h)=3(x+h)^2+2(x+h)-1=3x^2+6xh+3h^2+2x+2h-1.\] \[f(x+h)-f(x)=6xh+3h^2+2h.\] \[\frac{f(x+h)-f(x)}{h}=6x+3h+2.\] \[f'(x)=\lim_{h\to 0}(6x+3h+2)=6x+2.\]

So \(\dfrac{d}{dx}(3x^2+2x-1)=6x+2\).

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Write each term using \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).

\[\binom{2x+1}{3}=\frac{(2x+1)(2x)(2x-1)}{6}=\frac{8x^3-2x}{6},\] \[\binom{2x-1}{3}=\frac{(2x-1)(2x-2)(2x-3)}{6}=\frac{8x^3-24x^2+22x-6}{6}.\]

Their difference:

\[\binom{2x+1}{3}-\binom{2x-1}{3}=\frac{(8x^3-2x)-(8x^3-24x^2+22x-6)}{6}=\frac{24x^2-24x+6}{6}=4x^2-4x+1.\]

And \(2\binom{x}{2}=2\cdot\dfrac{x(x-1)}{2}=x^2-x\). Therefore

\[\binom{2x+1}{3}-\binom{2x-1}{3}-2\binom{x}{2}=(4x^2-4x+1)-(x^2-x)=3x^2-3x+1.\]

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(a) For parallelogram \(PQRS\) the diagonals bisect each other, so midpoint of \(PR\) = midpoint of \(QS\).

\[\left(\frac{-1+x}{2},\frac{4+y}{2}\right)=\left(\frac{2+(-2)}{2},\frac{3+3}{2}\right)=(0,3).\] \[\frac{-1+x}{2}=0\Rightarrow x=1,\qquad \frac{4+y}{2}=3\Rightarrow y=2.\]

So \(R(1,2)\), i.e. \(x=1,\ y=2\). (Check: \(\vec{PQ}=(3,-1)=\vec{SR}\).)

(b) From rest, \(u=0\), reaches \(v=20\ \text{m/s}\) after \(s=8\ \text{m}\).

(i) Acceleration: \(v^2=u^2+2as\Rightarrow 20^2=2a(8)\Rightarrow a=\dfrac{400}{16}=25\ \text{m/s}^2\).

(ii) Time to travel 40 m: \(s=ut+\tfrac12 at^2\Rightarrow 40=\tfrac12(25)t^2=12.5t^2\).

\[t^2=3.2\Rightarrow t=\sqrt{3.2}\approx 1.79\ \text{s}.\]

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(a) \(\sin p=\tfrac12\) with \(p\) acute gives \(p=30^\circ\), so \(\cos p=\tfrac{\sqrt3}{2}\). Since \(90^\circ \[\sin(p-q)=\sin p\cos q-\cos p\sin q=\frac12\!\left(-\frac13\right)-\frac{\sqrt3}{2}\cdot\frac{2\sqrt2}{3}=-\frac16-\frac{\sqrt6}{3}=-\frac{1+2\sqrt6}{6}\approx -0.98.\]

(b) Trapezium rule, seven ordinates \(\Rightarrow\) six strips, \(h=\dfrac{4-1}{6}=0.5\), with \(f(x)=\dfrac{2}{\sqrt{x+3}}\):

x	1	1.5	2	2.5	3	3.5	4
f(x)	1.0000	0.9428	0.8944	0.8528	0.8165	0.7845	0.7559

\[\int_1^4 f\,dx\approx\frac{h}{2}\Big[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)\Big]=\frac{0.5}{2}\big[1.7559+2(4.2910)\big].\] \[=0.25(1.7559+8.5820)=0.25(10.3379)\approx 2.58.\]

(The exact value \(\big[4\sqrt{x+3}\big]_1^4=4(\sqrt7-2)=2.58\) confirms this.)

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(a) For external division of \((x_1,y_1)=(7,-5)\) and \((x_2,y_2)=(-2,7)\) in the ratio \(m:n = 3:2\):

\[x = \frac{m x_2 - n x_1}{m - n} = \frac{3(-2) - 2(7)}{3 - 2} = \frac{-6 - 14}{1} = -20\] \[y = \frac{m y_2 - n y_1}{m - n} = \frac{3(7) - 2(-5)}{3 - 2} = \frac{21 + 10}{1} = 31\]

The point is \((-20,\ 31)\).

(b) Rationalise each term.

\[\frac{2}{1+\sqrt2}\times\frac{\sqrt2-1}{\sqrt2-1} = \frac{2(\sqrt2-1)}{(\sqrt2)^2-1^2}=\frac{2(\sqrt2-1)}{1}=2\sqrt2-2\] \[\frac{2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2} = \frac{2(2-\sqrt2)}{4-2}=\frac{2(2-\sqrt2)}{2}=2-\sqrt2\]

Therefore

\[\frac{2}{1+\sqrt2}-\frac{2}{2+\sqrt2} = (2\sqrt2-2)-(2-\sqrt2) = 3\sqrt2 - 4\]

In the form \(p+q\sqrt n\): \(p=-4,\ q=3,\ n=2\), i.e. \(-4+3\sqrt2\).

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(a) Taking \(m=i-j\) and \(n=2i+3j\), with \(2m+n-r=0\) we get \(r=2m+n\).

\[r=2(i-j)+(2i+3j)=(2i-2j)+(2i+3j)=4i+j\]

\[|r|=\sqrt{4^{2}+1^{2}}=\sqrt{17}\approx4.12\]

(b) \(S=3t-\dfrac{t^{3}}{3}+9\). Velocity is \(v=\dfrac{dS}{dt}=3-t^{2}\).

(i) Time when momentarily at rest: \(v=0\):

\[3-t^{2}=0\ \Rightarrow\ t^{2}=3\ \Rightarrow\ t=\sqrt{3}\ \text{s}\ (\approx1.73\ \text{s})\]

(ii) Distance at that instant:

\[S=3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3}+9=3\sqrt{3}-\sqrt{3}+9=2\sqrt{3}+9\approx12.46\ \text{m}\]

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Hit probabilities: \(P(X)=\tfrac13,\ P(Y)=\tfrac15,\ P(Z)=\tfrac14\); miss probabilities: \(\tfrac23,\tfrac45,\tfrac34\). "Only one hits" means exactly one succeeds while the other two miss (events independent):

\[P(\text{only }X)=\frac13\cdot\frac45\cdot\frac34=\frac{12}{60}=\frac15,\] \[P(\text{only }Y)=\frac23\cdot\frac15\cdot\frac34=\frac{6}{60}=\frac{1}{10},\] \[P(\text{only }Z)=\frac23\cdot\frac45\cdot\frac14=\frac{8}{60}=\frac{2}{15}.\]

Adding (LCD 30):

\[P(\text{exactly one})=\frac{6}{30}+\frac{3}{30}+\frac{4}{30}=\frac{13}{30}\approx 0.43.\]

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With two fair dice there are \(36\) equally likely outcomes.

Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) gives \(6\) outcomes, so \(P(7) = \dfrac{6}{36} = \dfrac{1}{6}\).

Sum of 4: \((1,3),(2,2),(3,1)\) gives \(3\) outcomes, so \(P(4) = \dfrac{3}{36} = \dfrac{1}{12}\).

The two throws are independent, so

\[P(\text{7 then 4}) = \frac{1}{6}\times\frac{1}{12} = \frac{1}{72}\]

Probability \(= \dfrac{1}{72}\).

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Given \(\sin x = \dfrac{P-Q}{P+Q}\) with \(0^{o}\le x\le 90^{o}\).

\[\cos^2 x = 1 - \sin^2 x = 1 - \frac{(P-Q)^2}{(P+Q)^2} = \frac{(P+Q)^2-(P-Q)^2}{(P+Q)^2} = \frac{4PQ}{(P+Q)^2}\] \[\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{(P-Q)^2/(P+Q)^2}{4PQ/(P+Q)^2} = \frac{(P-Q)^2}{4PQ}\]

Hence

\[1 - \tan^2 x = 1 - \frac{(P-Q)^2}{4PQ} = \frac{4PQ - (P^2 - 2PQ + Q^2)}{4PQ} = \frac{6PQ - P^2 - Q^2}{4PQ}\]

\(\displaystyle 1 - \tan^2 x = \frac{6PQ - P^2 - Q^2}{4PQ}\).

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(a) The line \(y + 2x - 3 = 0\) gives \(y = 3 - 2x\). At intersection with \(y = 7 - \dfrac{6}{x}\):

\[7 - \frac{6}{x} = 3 - 2x\]

Multiply through by \(x\):

\[7x - 6 = 3x - 2x^2 \;\Rightarrow\; 2x^2 + 4x - 6 = 0 \;\Rightarrow\; x^2 + 2x - 3 = 0\] \[(x+3)(x-1)=0 \;\Rightarrow\; x = -3 \text{ or } x = 1\]

When \(x = 1:\ y = 3 - 2 = 1\). When \(x = -3:\ y = 3 + 6 = 9\).

The points are \((1,\ 1)\) and \((-3,\ 9)\).

(b) Midpoint \(= \left(\dfrac{1 + (-3)}{2},\ \dfrac{1 + 9}{2}\right) = (-1,\ 5)\).

Gradient of the join \(= \dfrac{9 - 1}{-3 - 1} = \dfrac{8}{-4} = -2\); the perpendicular gradient is \(\dfrac{1}{2}\).

\[y - 5 = \tfrac{1}{2}(x + 1) \;\Rightarrow\; 2y - 10 = x + 1 \;\Rightarrow\; x - 2y + 11 = 0\]

Perpendicular bisector: \(x - 2y + 11 = 0\).

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(a) For \(4x^2 - px + 1 = 0\) to have real roots, the discriminant must be non-negative:

\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-p)^2 - 4(4)(1) \ge 0 \;\Rightarrow\; p^2 - 16 \ge 0\] \[(p-4)(p+4) \ge 0 \;\Rightarrow\; p \le -4 \ \text{ or } \ p \ge 4\]

(b)(i) Using \((1+3x)^6 = \displaystyle\sum_{r=0}^{6}\binom{6}{r}(3x)^r\):

\[(1+3x)^6 = 1 + 18x + 135x^2 + 540x^3 + 1215x^4 + 1458x^5 + 729x^6\]

(ii) Put \(3x = 0.03\), i.e. \(x = 0.01\), so \((1+3x)^6 = (1.03)^6\):

\[(1.03)^6 \approx 1 + 18(0.01) + 135(0.01)^2 + 540(0.01)^3 + \dots\] \[= 1 + 0.18 + 0.0135 + 0.00054 + 0.0000122 + \dots = 1.19405\]

To four significant figures, \((1.03)^6 \approx \mathbf{1.194}\).

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The data provided cover 9 students (A to I), so \(n = 9\). Rank each set with 1 for the highest score.

Student	Entrance	Rank \(R_x\)	School	Rank \(R_y\)	\(d = R_x - R_y\)	\(d^2\)
A	11	6	5	8	-2	4
B	12	5	10	3	2	4
C	8	8	9	4	4	16
D	13	4	7	6	-2	4
E	6	9	4	9	0	0
F	15	2	8	5	-3	9
G	10	7	6	7	0	0
H	14	3	14	1	2	4
I	17	1	11	2	-1	1
Total \(\sum d^2\)						42

1. Spearman's rank correlation coefficient.

\[ r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 42}{9(81 - 1)} = 1 - \frac{252}{720} = 1 - 0.35 = \mathbf{0.65} \]

2. Conclusion. \(r_s = 0.65\) is a fairly strong positive correlation. Students who perform well in the entrance examination tend also to perform well subsequently in school, so the researcher would conclude that entrance-examination performance is a reasonably good indicator of later school performance.

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System: \(5x-6y+4z=15,\ 7x+4y-3z=19,\ 2x+y+6z=46\). Use Cramer's rule.

\[D=\begin{vmatrix}5&-6&4\\7&4&-3\\2&1&6\end{vmatrix}=5(24+3)+6(42+6)+4(7-8)=135+288-4=419.\] \[D_x=\begin{vmatrix}15&-6&4\\19&4&-3\\46&1&6\end{vmatrix}=15(27)+6(252)+4(-165)=405+1512-660=1257.\] \[D_y=\begin{vmatrix}5&15&4\\7&19&-3\\2&46&6\end{vmatrix}=5(252)-15(48)+4(284)=1260-720+1136=1676.\] \[D_z=\begin{vmatrix}5&-6&15\\7&4&19\\2&1&46\end{vmatrix}=5(165)+6(284)+15(-1)=825+1704-15=2514.\] \[x=\frac{D_x}{D}=\frac{1257}{419}=3,\quad y=\frac{D_y}{D}=\frac{1676}{419}=4,\quad z=\frac{D_z}{D}=\frac{2514}{419}=6.\]

So \(x=3,\ y=4,\ z=6\). Check in equation 2: \(7(3)+4(4)-3(6)=21+16-18=19\).

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For \(3x^2+4x-5=0\) with roots \(\alpha,\beta\): sum \(\alpha+\beta=-\dfrac{4}{3}\), product \(\alpha\beta=-\dfrac{5}{3}\).

Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\):

\[(\alpha-\beta)^2=\left(-\frac43\right)^2-4\left(-\frac53\right)=\frac{16}{9}+\frac{20}{3}=\frac{16}{9}+\frac{60}{9}=\frac{76}{9}.\] \[\alpha-\beta=\sqrt{\frac{76}{9}}=\frac{\sqrt{76}}{3}=\frac{2\sqrt{19}}{3}.\]

So \(\alpha-\beta=\dfrac{2\sqrt{19}}{3}\) (taking the positive surd).

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The 12 kg mass hangs from a string whose upper end is fixed at \(P\). A horizontal force \(R\) pulls the mass sideways so that the string makes \(45^{\circ}\) with the vertical. Three forces act at the junction (knot): the tension \(T\) along the string, the horizontal pull \(R\), and the weight \(W\) acting vertically downward.

Weight of the mass (taking \(g = 10\,\text{m s}^{-2}\)):

\[ W = mg = 12 \times 10 = 120\,\text{N} \]

Since the system is in equilibrium, resolve the tension into vertical and horizontal components. The string is \(45^{\circ}\) from the vertical, so the vertical component of \(T\) is \(T\cos 45^{\circ}\) and the horizontal component is \(T\sin 45^{\circ}\).

(a) Tension in the string

Resolving vertically (the vertical component of the tension supports the weight):

\[ T\cos 45^{\circ} = W \]\[ T \times \frac{\sqrt{2}}{2} = 120 \]\[ T = \frac{120}{\cos 45^{\circ}} = \frac{120}{0.7071} = 120\sqrt{2} \]\[ T \approx 169.7\,\text{N} \]

So the tension is about \(170\,\text{N}\).

(b) Value of \(R\)

Resolving horizontally (the horizontal force balances the horizontal component of the tension):

\[ R = T\sin 45^{\circ} = 120\sqrt{2} \times \frac{\sqrt{2}}{2} \]\[ R = 120\sqrt{2} \times 0.7071 = 120\,\text{N} \]

Hence \(R = 120\,\text{N}\).

Check: Because the string is at \(45^{\circ}\), the vertical and horizontal components of \(T\) are equal, so \(R\) equals the weight, \(120\,\text{N}\), which agrees with the result above. (Using \(g = 9.8\,\text{m s}^{-2}\) gives \(W = 117.6\,\text{N}\), \(T \approx 166.3\,\text{N}\) and \(R = 117.6\,\text{N}\).)

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(a) Reading the images off column by column:

\[M=\begin{pmatrix}7&0\\3&-1\end{pmatrix},\qquad N=\begin{pmatrix}2&-1\\5&3\end{pmatrix}\]

(b) Image of \(P(2,-3)\) under N then M.

Apply \(N\): \((2x-y,\ 5x+3y)=(2(2)-(-3),\ 5(2)+3(-3))=(7,\ 1)\).

Apply \(M\) to \((7,1)\): \((7x,\ 3x-y)=(7(7),\ 3(7)-1)=(49,\ 20)\).

Image is \((49,\ 20)\).

(c) Point whose image under N is \((2,4)\). Solve \(N(x,y)=(2,4)\):

\[2x-y=2,\qquad 5x+3y=4\]

From the first, \(y=2x-2\). Substituting: \(5x+3(2x-2)=4\Rightarrow11x-6=4\Rightarrow x=\dfrac{10}{11}\).

\[y=2\left(\tfrac{10}{11}\right)-2=-\tfrac{2}{11}\]

The point is \(\left(\dfrac{10}{11},\ -\dfrac{2}{11}\right)\).

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First find the y-intercept of \(3x-2y+6=0\): put \(x=0\Rightarrow -2y+6=0\Rightarrow y=3\). The point is \((0,3)\).

The radius is the distance from the centre \((2,3)\) to \((0,3)\):

\[r=\sqrt{(2-0)^2+(3-3)^2}=\sqrt{4}=2.\]

Equation of the circle:

\[(x-2)^2+(y-3)^2=2^2\Rightarrow (x-2)^2+(y-3)^2=4,\]

or \(x^2+y^2-4x-6y+9=0\).

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Method. Assumed mean \(A = 13\), class width \(c = 5\), \(u = \dfrac{x - 13}{5}\) where \(x\) is the class midpoint.

Age (years)	Midpoint \(x\)	\(u\)	\(f\)	\(fu\)
1 - 5	3	-2	18	-36
6 - 10	8	-1	12	-12
11 - 15	13	0	25	0
16 - 20	18	1	15	15
21 - 25	23	2	20	40
26 - 30	28	3	10	30
Total			100	37

Mean age.

\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 13 + \frac{37}{100}\times 5 = 13 + 1.85 = \mathbf{14.85 \text{ years}} \]

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The series \(-3,\ -1,\ 1,\ \dots\) is arithmetic with first term \(a=-3\) and common difference \(d=2\).

\[S_n = \frac{n}{2}\big[2a + (n-1)d\big] = \frac{n}{2}\big[-6 + 2(n-1)\big] = \frac{n}{2}(2n-8) = n(n-4)\]

Set \(S_n = 165\):

\[n(n-4) = 165 \;\Rightarrow\; n^2 - 4n - 165 = 0\] \[n = \frac{4 \pm \sqrt{16 + 660}}{2} = \frac{4 \pm \sqrt{676}}{2} = \frac{4 \pm 26}{2}\]

Taking the positive value, \(n = \dfrac{30}{2} = 15\).

15 terms are required.

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Write every base as a power of \(5\): \(625=5^{4}\), \(125=5^{3}\).

Numerator:

\[625^{\left(\frac{3x}{4}-1\right)}=5^{4\left(\frac{3x}{4}-1\right)}=5^{3x-4},\qquad 125^{(x-1)}=5^{3(x-1)}=5^{3x-3}\]

So the expression is

\[\frac{5^{3x-4}+5^{3x-3}}{5^{3x-2}}=\frac{5^{3x-4}}{5^{3x-2}}+\frac{5^{3x-3}}{5^{3x-2}}=5^{-2}+5^{-1}\]

\[=\frac{1}{25}+\frac{1}{5}=\frac{1}{25}+\frac{5}{25}=\frac{6}{25}\]

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Measure each direction anticlockwise from the positive x-axis and resolve. For equilibrium the sum of the components in each direction is zero.

x-components:

\[5\cos30^{o} + P\cos60^{o} + Q\cos150^{o} + 3\cos180^{o} + 5\cos270^{o} = 0\] \[\tfrac{5\sqrt3}{2} + \tfrac{P}{2} - \tfrac{\sqrt3}{2}Q - 3 + 0 = 0\]

y-components:

\[5\sin30^{o} + P\sin60^{o} + Q\sin150^{o} + 3\sin180^{o} + 5\sin270^{o} = 0\] \[\tfrac{5}{2} + \tfrac{\sqrt3}{2}P + \tfrac{1}{2}Q + 0 - 5 = 0\]

The y-equation gives \(\sqrt3\,P + Q = 5\), so \(Q = 5 - \sqrt3\,P\).

The x-equation gives \(P - \sqrt3\,Q = 6 - 5\sqrt3\). Substituting:

\[P - \sqrt3(5 - \sqrt3 P) = 6 - 5\sqrt3 \;\Rightarrow\; 4P - 5\sqrt3 = 6 - 5\sqrt3 \;\Rightarrow\; 4P = 6\]

So \(P = 1.5\), and \(Q = 5 - 1.5\sqrt3 = 5 - 2.598 = 2.402\).

\(P \approx 1.5\,\text{N}, \quad Q \approx 2.4\,\text{N}\) (to 1 d.p.).

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Method (assumed mean / coding). Take assumed mean \(A = 67\) and class width \(c = 5\). For each class let \(x\) be the midpoint and \(u = \dfrac{x - A}{c}\).

Marks	Midpoint \(x\)	\(u\)	\(f\)	\(fu\)	\(fu^2\)
50 - 54	52	-3	5	-15	45
55 - 59	57	-2	15	-30	60
60 - 64	62	-1	20	-20	20
65 - 69	67	0	28	0	0
70 - 74	72	1	12	12	12
75 - 79	77	2	9	18	36
80 - 84	82	3	7	21	63
85 - 89	87	4	4	16	64
Total			100	2	300

(a) Mean.

\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 67 + \frac{2}{100}\times 5 = 67 + 0.1 = \mathbf{67.1} \]

(b) Standard deviation.

\[ \text{SD} = c\sqrt{\frac{\sum fu^2}{\sum f} - \left(\frac{\sum fu}{\sum f}\right)^2} = 5\sqrt{\frac{300}{100} - \left(\frac{2}{100}\right)^2} \] \[ = 5\sqrt{3 - 0.0004} = 5\sqrt{2.9996} = 5 \times 1.7319 \approx \mathbf{8.7} \]

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The three forces are \(10\,\text{N},\ 14\,\text{N}\) and \(16\,\text{N}\) in equilibrium, so the resultant of the \(10\,\text{N}\) and \(16\,\text{N}\) forces must be equal in magnitude to \(14\,\text{N}\) (and opposite to it).

Let \(\theta\) be the angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces. Using the resultant (parallelogram) formula:

\[R^2 = 10^2 + 16^2 + 2(10)(16)\cos\theta = 14^2\] \[100 + 256 + 320\cos\theta = 196\] \[320\cos\theta = 196 - 356 = -160 \;\Rightarrow\; \cos\theta = -\tfrac{1}{2}\] \[\theta = 120^{o}\]

The angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces is \(120^{o}\).

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Set distances from end \(W\). The beam is \(90\,\text{cm}\) long.

• \(Y\): \(|WY| = 10\,\text{cm}\), so \(Y\) is at \(10\,\text{cm}\).
• Pivot: \(35\,\text{cm}\) from \(W\).
• Weight \(50\,\text{N}\) acts at the centre (uniform beam): \(45\,\text{cm}\).
• \(Z\): \(|XZ| = 10\,\text{cm}\), so \(Z\) is at \(90 - 10 = 80\,\text{cm}\).

Taking moments about the pivot (which removes the pivot reaction). Distances from the pivot:

• \(T\) at \(Y\): \(35 - 10 = 25\,\text{cm}\) on the \(W\) side.
• Weight \(50\,\text{N}\): \(45 - 35 = 10\,\text{cm}\) on the \(X\) side.
• \(20\,\text{N}\) at \(Z\): \(80 - 35 = 45\,\text{cm}\) on the \(X\) side.

For equilibrium, the anticlockwise moment of \(T\) balances the clockwise moments of the weight and the \(20\,\text{N}\) force:

\[T \times 25 = 50 \times 10 + 20 \times 45\] \[25\,T = 500 + 900 = 1400\] \[T = \frac{1400}{25} = 56\,\text{N}\]

\(T = 56\,\text{N}\).

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This is a binomial situation. Let \(X\) be the number who fail. A candidate fails with probability \(p=0.4\) and passes with probability \(0.6\), with \(n=10\).

\[P(X=r)=\binom{10}{r}(0.4)^{r}(0.6)^{10-r}\]

(1) At least two failed: \(P(X\ge 2)\)

\[P(X\ge2)=1-P(0)-P(1)\]

\(P(0)=(0.6)^{10}=0.00605\) and \(P(1)=\binom{10}{1}(0.4)(0.6)^{9}=10(0.4)(0.010078)=0.04031\).

\[P(X\ge2)=1-0.00605-0.04031=0.954\]

(2) Exactly half passed means \(5\) passed and \(5\) failed, i.e. \(X=5\).

\[P(X=5)=\binom{10}{5}(0.4)^{5}(0.6)^{5}=252(0.01024)(0.07776)=0.201\]

(3) At most two failed: \(P(X\le 2)=P(0)+P(1)+P(2)\)

\(P(2)=\binom{10}{2}(0.4)^{2}(0.6)^{8}=45(0.16)(0.016796)=0.12093\).

\[P(X\le2)=0.00605+0.04031+0.12093=0.167\]

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(a) \(P(\text{bad})=0.3,\ P(\text{good})=0.7,\ n=12\), \(P(X=r)=\binom{12}{r}(0.3)^r(0.7)^{12-r}\) (\(X\)=number bad).

(i) exactly 6 bad: \(\binom{12}{6}(0.3)^6(0.7)^6=924(0.000729)(0.117649)=0.0793\).

(ii) at least 4 bad \(=1-\big[P(0)+P(1)+P(2)+P(3)\big]\):

\[P(0)=0.0138,\ P(1)=0.0712,\ P(2)=0.1678,\ P(3)=0.2397;\ \text{sum}=0.4925.\] \[P(X\ge 4)=1-0.4925=0.5075.\]

(iii) no bad one: \(P(0)=(0.7)^{12}=0.0138\).

(b) 8 boys, 5 girls; committee of 7 with 4 boys and 3 girls:

\[P=\frac{\binom{8}{4}\binom{5}{3}}{\binom{13}{7}}=\frac{70\times 10}{1716}=\frac{700}{1716}=\frac{175}{429}\approx 0.4079.\]

Awọn alaye Idahun

Awọn alaye Idahun

\(\vec{OP}=\begin{pmatrix}-3\\-4\end{pmatrix},\ \vec{OQ}=\begin{pmatrix}8\\-15\end{pmatrix}\).

Dot product: \(\vec{OP}\cdot\vec{OQ}=(-3)(8)+(-4)(-15)=-24+60=36\).

Magnitudes: \(|\vec{OP}|=\sqrt{9+16}=5\), \(|\vec{OQ}|=\sqrt{64+225}=\sqrt{289}=17\).

\[\cos\theta=\frac{36}{5\times 17}=\frac{36}{85}=0.4235\Rightarrow \theta=\cos^{-1}(0.4235)\approx 65^\circ.\]

The angle between the vectors is about \(65^\circ\).