WAEC SSCE - Further Mathematics - 2009 (Essay)
Ibeere 1 Ìròyìn
The probabilities of Rotey obtaining the highest mark in Mathematics, Physics and Biology tests are 0.9, 0.75 and 0.8 respectively. Calculate the probability of getting the highest marks in at least two of the subjects.
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 \(= 0.135 + 0.180 + 0.060 = 0.375\).
\[P(\text{at least two}) = 0.375 + 0.54 = \mathbf{0.915}\]Awọn alaye Idahun
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 \(= 0.135 + 0.180 + 0.060 = 0.375\).
\[P(\text{at least two}) = 0.375 + 0.54 = \mathbf{0.915}\]Ibeere 2 Ìròyìn
(a) The 3rd and 6th terms of a Geometric Progression (G.P) are 2 and 54 respectively. Find the : (i) common ratio ; (ii) first term ; (iii) sum of the first 10 terms, correct to the nearest whole number.
(b) The ratio of the coefficient of \(x^{4}\) to that of \(x^{3}\) in the binomial expansion of \((1 + 2x)^{n}\) is \(3 : 1\). Find the value of n.
(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 \(\dfrac{3}{1}\):
\[\frac{n - 3}{2} = 3 \;\Rightarrow\; n - 3 = 6 \;\Rightarrow\; n = 9\]Awọn alaye Idahun
(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 \(\dfrac{3}{1}\):
\[\frac{n - 3}{2} = 3 \;\Rightarrow\; n - 3 = 6 \;\Rightarrow\; n = 9\]Ibeere 3 Ìròyìn
(a) Solve : \(2x^{2} + x - 6 < 0\)
(b) Express \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\) in the form \(m\sqrt{2} + n\sqrt{5}\) where m and n are rational numbers.
(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 \(n = \dfrac{19}{43}\).
Awọn alaye Idahun
(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 \(n = \dfrac{19}{43}\).
Ibeere 4 Ìròyìn
(a) Using the same axes, sketch the curves \(y = 6 - x - x^{2}\) and \(y = 3x^{2} - 2x + 3\).
(b) Find the x- coordinates of the points of intersection of the two curves in (a).
(c) Calculatethe area of the finite region bounded by the two curves in (a).
(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.
Awọn alaye Idahun
(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.
Ibeere 5 Ìròyìn
(a) Use the trapezium rule with five ordinates to evaluate \(\int_{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\), correct to four significant figures.
(b) If \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\), find the image of the point (1, 2) under the linear transformation \(A^{2} + A + 2I\), where I is the \(2 \times 2\) unit matrix.
(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 |
(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 \((21,\ 37)\).
Awọn alaye Idahun
(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 |
(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 \((21,\ 37)\).
Ibeere 6 Ìròyìn
(a) The position vectors of points L and M are (5i + 6j) and (13i + 4j) respectively. If point K lies on LM such that LK : KM is 2 : 3, find the position vector of K.
(b) Three poles are situated at points A, B and C on the same horizontal plane such that \(AB = (8km, 060°)\) and \(BC = (12km, 130°)\). Calculate,
(i) |AC|, correct to three significant figures ; (ii) the bearing of C from A, correct to the nearest degree.
(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).
Awọn alaye Idahun
(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).
Ibeere 7 Ìròyìn
The table below shows the corresponding values of two variables X and Y.
| X | 33 | 31 | 28 | 25 | 23 | 22 | 19 | 17 | 16 | 14 |
| Y | 4 | 6 | 4 | 10 | 12 | 10 | 14 | 15 | 18 | 22 |
(a) Plot a scatter diagram to represent the data.
(b) Calculate \(\bar{x}\), the mean of X and \(\bar{y}\), the mean of Y.
(c) Draw the line of best fit to pass through \((\bar{x}, \bar{y})\).
(d) From your graph in (c), determine the (i) relationship between X and Y ; (ii) value of Y when X is 24.
(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 |
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).
Awọn alaye Idahun
(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 |
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).
Ibeere 8 Ìròyìn
(a) Evaluate \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\)
(b)(i) Evaluate: \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}\)
(ii) Using your answer in b(i), solve the simultaneous equations :
\(2x - 3y + z = 10\)
\(y - 2z = -7\)
\(x + 2y - 3z = -9\)
(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 \((-1) - 2(3) = -7\).
Awọn alaye Idahun
(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 \((-1) - 2(3) = -7\).
Ibeere 9 Ìròyìn
If the quadratic equation \((x + 1)(x + 2) = k(3x + 7)\) has equal roots, find the possible values of the constant k.
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 \(k = -1\).
Awọn alaye Idahun
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 \(k = -1\).
Ibeere 10 Ìròyìn
Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving your answer in surd form.
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:
\[\tan15^{o} = 2 - \sqrt3\]Awọn alaye Idahun
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:
\[\tan15^{o} = 2 - \sqrt3\]Ibeere 11 Ìròyìn
Solve the simultaneous equations : \(\log_{2} x - \log_{2} y = 2 ; \log_{2} (x - 2y) = 3\)
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 \(\log_2(16 - 8) = \log_2 8 = 3\).
Awọn alaye Idahun
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 \(\log_2(16 - 8) = \log_2 8 = 3\).
Ibeere 12 Ìròyìn
A student representative council consists of 8 girls and 6 boys. If an editorial board consisting of 5 persons is to be formed, what is the probability that the board consists of
(a) 3 girls and 2 boys ;
(b) either all girls or all boys.
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:
\[\binom{8}{5} + \binom{6}{5} = 56 + 6 = 62\] \[P = \frac{62}{2002} = \frac{31}{1001} \approx 0.031\]Awọn alaye Idahun
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:
\[\binom{8}{5} + \binom{6}{5} = 56 + 6 = 62\] \[P = \frac{62}{2002} = \frac{31}{1001} \approx 0.031\]Ibeere 13 Ìròyìn
(a) Simplify \(^{n + 1}C_{3} - ^{n - 1}C_{3}\)
(b) A fair die is thrown five times. Calculate, correct to three decimal places, the probability of obtaining (i) at most two sixes ; (ii) exactly three sixes.
(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:
\[P(3) = \binom{5}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^2 = 10\cdot\frac{1}{216}\cdot\frac{25}{36} = \frac{250}{7776} \approx 0.032\]Awọn alaye Idahun
(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:
\[P(3) = \binom{5}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^2 = 10\cdot\frac{1}{216}\cdot\frac{25}{36} = \frac{250}{7776} \approx 0.032\]Ibeere 14 Ìròyìn
The coordinates of the vertices of triangle ABC are A(-2, 1), B(4, -2) and C(1, 8) respectively. If D(x, y) is the foot perpendicular from A to BC, find
(a) an equation connecting x and y ;
(b) the unit vector in the direction of BC.
\(A(-2,1),\ B(4,-2),\ C(1,8)\). \(D(x,y)\) 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 \(BC\):
\[\hat{BC} = \frac{1}{\sqrt{109}}\begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -\tfrac{3}{\sqrt{109}} \\[4pt] \tfrac{10}{\sqrt{109}} \end{pmatrix}\]Awọn alaye Idahun
\(A(-2,1),\ B(4,-2),\ C(1,8)\). \(D(x,y)\) 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 \(BC\):
\[\hat{BC} = \frac{1}{\sqrt{109}}\begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -\tfrac{3}{\sqrt{109}} \\[4pt] \tfrac{10}{\sqrt{109}} \end{pmatrix}\]Ibeere 15 Ìròyìn
(a) The position vectors of points A, B and C are \(i + 5j , 3i + 9j\) and \(-i + j\) respectively. (i) Show that points A, B and C are collinear; (ii) Determine the ratio \(|AB| : |BC|\).
(b) A uniform beam XY of mass 10 kg and length 24m is hunged horizontally from a cross bar by teo vertical inextensible strings, one attached to X and the other at a point M, 4m away from Y. A mass of 50kg is suspended at a point N which is 8m from X. If the system remains in equilibrium, calculate the tensions in the strings.
(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 \(260\text{ N}\).
Awọn alaye Idahun
(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 \(260\text{ N}\).
Ibeere 16 Ìròyìn
Coplanar force 4N, 8N, 6N, 4N and 5N act at a point as shown in the diagram. If the 6N force acts in the direction 090°, calculate the :
(a) magnitude of the resultant force;
(b) direction of the resultant force.
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 \(090^\circ-36.2^\circ\approx \mathbf{054^\circ}\).
Awọn alaye Idahun
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 \(090^\circ-36.2^\circ\approx \mathbf{054^\circ}\).
Ibeere 17 Ìròyìn
The position vector of a body, with respect to the origin, is given by \(r = 4ti + (12 - 3t)j\) at any time t seconds.
(a) Find the velocity of the body ;
(b) Calculate the magnitude of the displacement between t = 0 and t = 5.
\(\mathbf{r} = 4t\,\mathbf{i} + (12 - 3t)\,\mathbf{j}\).
(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 \(= \mathbf{r}(5) - \mathbf{r}(0) = 20\,\mathbf{i} - 15\,\mathbf{j}\).
\[|\text{displacement}| = \sqrt{20^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25\ \text{units}\]Awọn alaye Idahun
\(\mathbf{r} = 4t\,\mathbf{i} + (12 - 3t)\,\mathbf{j}\).
(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 \(= \mathbf{r}(5) - \mathbf{r}(0) = 20\,\mathbf{i} - 15\,\mathbf{j}\).
\[|\text{displacement}| = \sqrt{20^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25\ \text{units}\]Ibeere 18 Ìròyìn
(a) The distribution of the lives (in days) of 40 transitor batteries is shown in the table:
| Battery life (in days) |
26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Frequency | 4 | 7 | 13 | 8 | 6 | 2 |
(a) Draw a histogram for the distribution.
(b) Use your graph in (a) to determine the mode for the distribution.
(c) Using an assumed mean of 43 days, calculate the mean of the distribution.
(d) What percentage number of batteries will live for less than 31 days or more than 45 days?
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.
Awọn alaye Idahun
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.
