Correct Answer: C. 0.879

Explanation

The Spearman's correlation coefficient \(\rho\) is given as:

\(\rho = 1 - \frac{6\sum d^{2}}{n(n^{2} - 1)}\)

= \(\rho = 1 - \frac{6 \times 20}{10(10^{2} - 1)}\)

= \(1 - \frac{120}{990} = \frac{870}{990}\)

= \(0.879\)

Explanation

(a) \(x = (-2i + 5j) ; y = (i - 7j)\)

(i) \((3x + 2y) = 3(-2i + 5j) + 2(i - 7j)\)

= \((-6i + 15j) + (2i - 14j)\)

= \(-4i + j\)

(ii) \((y - 2x) = (i - 7j) - 2(-2i + 5j)\)

\((i - 7j) - (-4i + 10j) = (5i - 17j)\)

\(|y - 2x| = |-3i - 17j| = \sqrt{(5^{2}) + (-17 ^{2})}\)

= \(\sqrt{25 + 289} = \sqrt{314}\)

(iii) Let the angle between x and y be \(\theta\).

\(x . y = |x||y| \cos \theta\)

\((-2i + 5j) . (i - 7j) = |-2i + 5j||i - 7j| \cos \theta\)

\(-2 - 35 = (\sqrt{(-2)^{2} + (5)^{2}})(\sqrt{(1)^{2} + (-7)^{2}}) \cos \theta\)

\(-37 = (\sqrt{29})(\sqrt{50}) \cos \theta \implies -37 = (\sqrt{1450})\cos \theta\)

\(\cos \theta = \frac{-37}{\sqrt{1450}} = -0.9717\)

\(\theta = \cos^{-1} (-0.9717) = 166.34°\)

(iv) \(x + y = (-2i + 5j) + (i - 7j) = -i - 2j\)

\(|x + y| = \sqrt{(-1)^{2} + (-2)^{2}} = \sqrt{5}\)

Unit vector = \(\frac{-i - 2j}{\sqrt{5}}\)

(b)

\(m_{1} = 0.084kg ; m_{2} = 20 kg ; m = 20.084 kg\)

\(u_{1} = ? ; u_{2} = 0 m/s ; v = 0.24 m/s\)

Momentum before collision = \(0.084 u_{1} kg m/s\)

Momentum after collision = \(20.084 \times 0.24 = 4.82016 kg m/s\)

\(\therefore 0.084 u_{1} = 4.82016 \)

\(u_{1} = \frac{4.82016}{0.084} = 57.3829 m/s\)

The initial velocity of bullet is 57.38 ms\(^{-1}\).

Correct Answer: B. 60°

Explanation

\(2\sin^{2} \theta = 1 + \cos \theta\)

\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)

\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)

\(2\cos^{2} \theta + \cos \theta - 1 = 0\)

Factorizing, we have

\((\cos \theta + 1)(2\cos \theta - 1) = 0\)

Note: In the range, \(0° \leq \theta \leq 90°\), all trig functions are positive, so we consider

\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)

\(\theta = 60°\).

Explanation

(a) \(\frac{\sqrt{5} + 4}{3 - 2\sqrt{5}} - \frac{2 + \sqrt{5}}{4 - 2\sqrt{5}}\)

= \(\frac{(\sqrt{5} + 4)(4 - 2\sqrt{5}) - (2 + \sqrt{5})(3 - 2\sqrt{5})}{(3 - 2\sqrt{5})(4 - 2\sqrt{5})}\)

= \(\frac{4\sqrt{5} - 10 + 16 - 8\sqrt{5} - 6 + 4\sqrt{5} - 3\sqrt{5} + 10}{12 - 6\sqrt{5} - 8\sqrt{5} + 4(5)}\)

= \(\frac{10 - 3\sqrt{5}}{32 - 14\sqrt{5}}\)

Rationalizing,

\((\frac{10 - 3\sqrt{5}}{32 - 14\sqrt{5}}) (\frac{32 + 14\sqrt{5}}{32 + 14\sqrt{5}})\)

= \(\frac{320 + 140\sqrt{5} - 96\sqrt{5} - 42(5)}{1024 + 448\sqrt{5} - 448\sqrt{5} - 196(5)}\)

= \(\frac{110 + 44\sqrt{5}}{44}\)

= \(\frac{5}{2} + \sqrt{5}\).

You can do the rationalizing before solving if you wish to. I just like mine a little complex.

(b)(i) \(\begin{pmatrix} 2 & -1 & 2 \\ 1 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix}\)

= \(2(3 - 8) - (-1)(1 - 4) + 2(2 - 3)\)

= \(2(-5) + 1(-3) + 2(-1)\)

= \(-10 - 3 - 2\)

= \(-15\)

(ii) \(2x - y + 2z = -5 ... (1)\)

\(x + 3y + 4z = 1 ..... (2)\)

\(x + 2y + z = -2 ..... (3)\)

\(x = \frac{\begin{vmatrix} -5 & -1 & 2 \\ 1 & 3 & 4 \\ -2 & 2 & 1 \end{vmatrix}}{-15}\)

= \(\frac{-5(3 - 8) + 1(1 + 8) + 2(2 + 6)}{-15}\)

= \(\frac{25 + 9 + 16}{-15}\)

= \(\frac{50}{-15} = -3.33\) (2 d.p)

Correct Answer: D. \(\frac{50 \sin \theta}{\sin (\theta + 45)°}\)

Explanation

\(^{5y}{C}_2\) = 190

\(\frac{5y(5y-1) (5y-2)!}{(5y-2)!(2)!}\) = 190

\(\frac{5y(5y-1)}{2}\) = 190

25y\(^2\) - 5y = 380

5y\(^2\) - 5y - 76 = 0

= (y-4) (5y+ 19) = 0

Either y - 4 = 0 or 5y+ 19 = 0;

y = 4 or -19/5

: y= 4

Explanation

\(3x^{2} + 5x + 1 = 0\)

\(x^{2} + \frac{5}{3}x + \frac{1}{3} = 0\)

\(\alpha \beta = \frac{c}{a}\)

= \(\frac{1}{3}\)

\(\alpha + \beta = \frac{-b}{a}\)

= \(-\frac{5}{3}\)

\((\alpha + \beta)^{3} = \alpha^{3} + 3\alpha^{2} \beta + 3\alpha \beta^{2} + \beta^{3}\)

\(\therefore \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha^{2} \beta - 3\alpha \beta^{2}\)

= \((\alpha + \beta)^{3} - 3\alpha \beta (\alpha + \beta)\)

= \((-\frac{5}{3})^{3} - 3(\frac{1}{3})(-\frac{5}{3})\)

= \((-\frac{125}{27}) + \frac{5}{3}\)

\(\therefore (\alpha^{3} + \beta^{3}) = -\frac{80}{27}\)

\(27(\alpha^{3} + \beta^{3}) = 27(-\frac{80}{27})\)

= -80.

Correct Answer: A. 13 metres

Explanation

\(v^{2} = u^{2} + 2as\)

\(v^{2} = u^{2} - 2gs\)

\(0 = 10^{2} - 2(10)s \implies -100 = -20s\)

\(s = 5 m + 8 m = 13m\) (8m is the height from where it was thrown)

Correct Answer: C. 0

Explanation

\(2\log_{4} 2 = x + 1\)

\(\log_{4} 2^{2} = \log_{4} 4 = 1\)

\(x + 1 = 1 \implies x = 0\)

Correct Answer: B. \(\frac{13}{54}\)

Explanation

P(only one hit target) = P(Kofi not Ama) + P(Ama not Kofi)

P(Kofi not Ama) = P(Kofi and Ama') = \(\frac{1}{6} \times \frac{8}{9} = \frac{8}{54}\)

P(Ama not Kofi) = P(Ama and Kofi') = \(\frac{1}{9} \times \frac{5}{6} = \frac{5}{54}\)

P(only one hit target) = \(\frac{8}{54} + \frac{5}{54} = \frac{13}{54}\)