Correct Answer: A. \(\frac{1}{4}\)

Explanation

Y = {18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29}

Number of prime numbers = 3

Prob(picking a prime number) = \(\frac{3}{12}\)

= \(\frac{1}{4}\)

Explanation

Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\)

= \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\)

= \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! (n - 3)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)!}{4! (n - 5)!}\)

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

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

= \(\frac{(n - 1)(n - 2) [n^{2} + n - n^{2} + 7n - 12]}{24}\)

= \(\frac{(n - 1)(n - 2)[8n - 12]}{24}\)

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

Correct Answer: C. 5

Explanation

\(\begin{array}{c|c} x & x - x & (x - \bar{x})^2\\ \hline 15 & -6 & 36\\21 & 0 & 0\\17 & -4 & 16\\ 26 & 5 & 25 \\ 18 & -3 &9 \\ 29 & 8 & 64 \end{array}\)

\(E(x - \bar{x})^2\) = 150

N = 6

S.D = \(\sqrt{\frac{(x - x)^2}{N}}\)

S.D = \(\sqrt{\frac{150}{6}}\) = 5

Correct Answer: B. 720mins

Explanation

(a) Let \(y = \frac{x^{2} + 1}{(x + 1)^{2}}\)

Let \(u = x^{2} + 1 ; v = (x + 1)^{2}\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x ; \frac{\mathrm d v}{\mathrm d x} = 2(x + 1)\)

Using the quotient rule,

\(\frac{\mathrm d y}{\mathrm d x} = \frac{v \frac{\mathrm d u}{\mathrm d x} - u \frac{\mathrm d v}{\mathrm d x}}{v^{2}}\)

= \(\frac{(x + 1)^{2} (2x) - (x^{2} + 1)(2(x + 1))}{((x + 1)^{2})^{2}}\)

= \(\frac{(x + 1)[(x + 1)(2x) - (x^{2} + 1)(2)}{(x + 1)^{4}}\)

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

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

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

= \(1(9 + 1) - 2(6 - 1) - 1(2 + 3)\)

= \(10 - 10 - 5 = -5\)

(ii) \(x + 2y - z = 4\)

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

\(-x + y + 3z = -1\)

In matrix form, this is

\(\begin{pmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

Cofactors :

\(A_{11} = +(3 \times 3 + 1) = 10\)

\(A_{12} = -(2 \times 3 - 1) = -5\)

\(A_{13} = +(2 \times 1 + 3) = 6\)

\(A_{21} = -(2 \times 3 + 1) = -7\)

\(A_{22} = +(1 \times 3 - 1) = 2\)

\(A_{23} = -(1 \times 1 + 1 \times 2) = -3\)

\(A_{31} = +(2 \times -1 + 3) = 1\)

\(A_{32} = -(-1 + 2) = -1\)

\(A_{33} = +(1 \times 3 - 2 \times 2) = -1\)

\(C = \begin{pmatrix} 10 & -5 & 5 \\ -7 & 2 & -3 \\ 1 & -1 & -1 \end{pmatrix}\)

adj A = \(C^{T} = \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\frac{1}{-5} \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix}\)

\(\therefore x = A^{-1} . b \)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

= \(\begin{pmatrix} - 8 + \frac{14}{5} + \frac{1}{5} \\ 4 - \frac{4}{5} + \frac{1}{5} \\ -4 + \frac{6}{5} - \frac{1}{5} \end{pmatrix}\)

= \(\begin{pmatrix} -5 \\ 3 \\ -3 \end{pmatrix}\)

x = -5 ; y = 3 ; z = -3.

Correct Answer: C. \(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 6\sqrt{10} - 2)\)

Explanation

\(\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}} = (\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}})(\frac{2 - 3\sqrt{2}}{2 - 3\sqrt{2}})\)

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{4 - 6\sqrt{2} + 6\sqrt{2} - 18}\)

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{-14}\)

= \(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 2 - 6\sqrt{10})\) (dividing through with the minus sign)

Correct Answer: C. 0.00360

Explanation

0,00 3597 = 0.00360 to 3 s.f

Correct Answer: C. 48kg 150g

Explanation

Since the class intervals are not equal, we draw the histogram in a way that the area of the bars, rather than the height must be proportional to the frequency. On the vertical scale, we plot frequency density instead of frequency where frequency density = frequency/ class width.

Correct Answer: C. 3xy³

Explanation

3\(\sqrt{27x^3y^9}\) = 3\(\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9}\)

= 3 \(\times x \times y^3\)

= 3xy³