Explanation

(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)

= \(\frac{1333}{268}\)

= \(4.974\) (to 3 d.p.)

(b) Even numbers = 147

p(even number) = \(\frac{147}{268}\)

= \(0.5485 \approxeq 0.549\)

(c) Odd numbers = 121

p(odd number) = \(\frac{121}{268}\)

= \(0.4515 \approxeq 0.452\)

Explanation

(i) \(T_{1} = \begin{pmatrix} -T_{1} \cos 40° \\ T_{1} \sin 40° \end{pmatrix} = \begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{2} \end{pmatrix}\)

\(T_{2} = \begin{pmatrix} T_{2} \cos 30° \\ T_{2} \cos 30° \end{pmatrix} = \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix}\)

\(W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}\)

(ii) P is at rest, the vector equation is

\(T_{1} + T_{2} + W = 0\)

(iii) \(\begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{1} \end{pmatrix} + \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix} + \begin{pmatrix} 0 \\ -65 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(0.7660T_{1} = 0.8660T_{2} ... (1)\)

\(0.6428T_{1} + 0.5000T_{2} - 65 = 0 ... (2)\)

From (1), \(T_{1} = \frac{0.8660T_{2}}{0.7660} ... (3)\)

Put (3) in (2) :

\(0.6428(\frac{0.8660T_{2}}{0.7660}) + 0.5000T_{2} = 65 \)

\(0.7627T_{2} + 0.5000T_{2} = 65\)

\(1.2267T_{2} = 65 \implies T_{2} = \frac{65}{1.2267}\)

\(T_{2} = 52.988N \approxeq 53.0N\) (to 1 d.p)

From (3), \(T_{1} = \frac{0.866 \times 52.988}{0.766} = 59.905N\)

\(\approxeq 59.9N\) (to 1 d.p)

Explanation

x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\)

Changing x and y to the form xi+yj

we have x= -4i + 3j and y = -9i - 15j

using cosØ = \(\frac{xy}{|x|ly|}\)

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(4\(^2\) +3\(^2\)) =√(16 +9) = √25 = 5

|y| = √(-9)\(^2\) + (-15)\(^2\) = √(81 +225) = √306

cosØ =\(\frac{-81}{5√306}\)

= \(\frac{-9√306}{170}\)

\(\frac{17.49 * 9}{170}\)

Ø = cos\(^{-1}\)(-0.1029);

Ø = 95.91°

Ø = 96°

Correct Answer: A. \(?26 units\)

Explanation

\(|MQ| = \frac{1}{2} |TQ|\)

\(|TQ| = √((y2 - y1)^2 + (x2 - x1)^2)\)

\(|TQ| = √((6 - 4)^2 + (-8 - 2)^2)\)

\(|TQ| = √(2^2 + (-10)^2)\)

\(|TQ| = √(4 + 100) = √104\)

\(|TQ| = 2√26 units\)

\(|MQ| = \frac{1}{2} |TQ| = 2 \times 2√26\)

∴ \(|MQ| = √26 units\)

Explanation

(a)(i) \(T_{3} = 2 ; ar^{2} = 2 ... (i)\)

\(T_{6} = 54 ; ar^{5} = 54 ... (ii)\)

\(\frac{T_{6}}{T_{3}} = \frac{ar^{5}}{ar^{2}} = \frac{54}{2}\)

\(r^{3} = 27 \implies r = 3 \)

(ii) First term = a

\(T_{3} = ar^{2} = 2\)

\(a = \frac{2}{3^{2}} = \frac{2}{9}\)

(iii) \(S_{10} = \frac{a(r^{n} - 1)}{r - 1}\)

= \(\frac{2(3^{10} - 1)}{9(3 - 1)}\)

= \(\frac{3^{10} - 1}{9}\)

= \(\frac{59049 - 1}{9} \approxeq 6561\) (nearest whole number).

(b) \((1 + 2x)^{n} = 1 + ^{n}C_{1} (1)^{n - 1} (2x) + ^{n}C_{2} (1)^{n - 2} (2x)^{2} + ^{n}C_{3} (1)^{n - 3} (2x)^{3} + ^{n}C_{4} (1)^{n - 4} (2x)^{4}+ ...\)

\(\frac{\text{coefficient of} x^{4}}{\text{coefficient of } x^{3}} = \frac{3}{1}\)

\(\frac{n!}{4! (n - 4)!} \times 2^{4} \times \frac{3! (n - 3)!}{n!} = \frac{3}{1}\)

\(\frac{16 \times 3! (n - 3)(n - 4)!}{4 \times 3! (n - 4)! \times 8} = \frac{3}{1}\)

\(\frac{2(n - 3)}{4} = \frac{3}{1} \implies 2n - 6 = 12\)

\(2n = 18 \implies n = 9\)

Correct Answer: D. 12

Explanation

Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)

Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)

\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)

\(r^{3} = \frac{972 \times 16}{9} = 1728 \implies r = \sqrt[3]{1728} = 12\)

Correct Answer: C. 336

Explanation

\(^3{P}_3\) = \(\frac{8!}{8-3}!\)

= \(\frac{(8 * 7 * 6 * 5)!}{5!}\)

= 336

Explanation

(a) Table of values for \(y = \sin 2x\) and \(y = \cos x\).

(b)(i) \(\sin 2x = 0 \implies x = 0°, 90°, 180°\)

(ii) \(\sin 2x - \cos x = 0 \implies x = 30°, 90°, 150°\).

Explanation

(a) \(a = \begin{pmatrix} 8 \\ 3 \end{pmatrix} ; b = \begin{pmatrix} 6 \\ -5 \end{pmatrix} ; c = \begin{pmatrix} 2 \\ -3 \end{pmatrix} ; |d| = \sqrt{41}\)

\(p = a + b - 2c\)

= \(\begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -3 \end{pmatrix}\)

= \(\begin{pmatrix} 8 + 6 + 4 \\ 3 - 5 - 6 \end{pmatrix}\)

= \(\begin{pmatrix} 10 \\ -8 \end{pmatrix}\)

\(|p| = \sqrt{10^{2} + (-8)^{2}} = \sqrt{164}\)

\(|p| = 2\sqrt{41}\)

\(|d| = \frac{1}{2} |p|\)

\(\therefore d = \frac{1}{2} p = \frac{1}{2} \begin{pmatrix} 10 \\ -8 \end{pmatrix}\)

= \(\begin{pmatrix} 5 \\ -4 \end{pmatrix}\)

(b) A(3, 4) ; B(3, n).

\(OA = 3i + 4j ; OB = 3i + nj\)

\(OA \cdot OB = |OA||OB| \cos \theta\)

\((3i + 4j) \cdot (3i + nj) = |3i + 4j||3i + nj| \cos 30\)

\(9 + 4n = 5 (\sqrt{9 + n^{2}}) (\sqrt{3}{2})\)

\(2(9 + 4n) = 5(\sqrt{3(9 + n^{2})})\)

\(18 + 8n = 5\sqrt{27 + 3n^{2}}\)

Squaring both sides,

\(324 + 288n + 64n^{2} = 25(27 + 3n^{2})\)

\(324 + 288n + 64n^{2} = 675 + 75n^{2}\)

\(75n^{2} - 64n^{2} - 288n + 675 - 324 = 0\)

\(11n^{2} - 288n + 351 = 0\)

\(n = \frac{-(-288) \pm \sqrt{(-288)^{2} - 4(11)(351)}}{2(11)}\)

\(n = \frac{288 \pm \sqrt{82944 - 15444}}{22}\)

\(n = \frac{288 \pm \sqrt{67500}}{22}\)

\(n = \frac{288 \pm 259.81}{22}\)

\(n = \frac{288 + 259.81}{22} ; n = \frac{288 - 259.81}{22}\)

\(n = \frac{547.81}{22} ; n = \frac{28.19}{22}\)

\(n = 24.9 ; n = 1.28\)

Correct Answer: B. \(225 - \frac{3x}{2}\)

Explanation

Let A and B be the sum for the boys and girls respectively.

\(\frac{A + B}{10 + 15} = \frac{A + B}{25} = 90\)

\(\implies A + B = 90 \times 25 = 2250\)

Given the average for girls = x, we have \(\frac{B}{15} = x \implies B = 15x)

\(\therefore A + 15x = 2250; A = 2250 - 15x \implies\) average score for boys \(= \frac{2250 - 15x}{10}\)

= \(225 - \frac{3x}{2}\)