Correct Answer: D. \( p \wedge q\)

Correct Answer: B. 0.30

Explanation

Let the number of students that like both Science and History = x

Number of Science only = 35 - x

Number of History only = 30 - x

35 - x + 30 - x + x = 50

65 - x = 50

x = 15

P(Science and History) = \(\frac{15}{50} = 0.30\)

Correct Answer: B. 4

Explanation

\(\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = 3(x - 2) - 2x = -2\)

\(3x - 6 - 2x = -2 \implies x - 6 = -2\)

\(x = -2 + 6 = 4\)

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

Explanation

P(even in 1 dice) = \(\frac{3}{6} = \frac{1}{2}\)

P(even in 2 fair die) = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

Correct Answer: A. 9

Explanation

\(f(x) = mx^{2} - 6x - 3\)

\(f '(x) = \frac{\mathrm d y}{\mathrm d x} = 2mx - 6\)

\(f'(1) = 2m - 6 = 12\)

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

Explanation

Let. X = Prob. that a good mango is selected = 18/24= 3/4

Y Prob. that a bad mango is selected = 6/24 1/4

Using the binomial probability distribution, we have:

(X+Y)\(^6\) = X\(^6\) + \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\) + \(^6{C}_4 {X}^2 {Y}^4\) + \(^6{C}_5 {X} {Y}^5\) + \(^6{C}_6 {Y}^6\)

Probability that not more than 3 are bad is

= \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\)

= 6(3/4)\(^5\) (1/4) + 15(3/4)\(^4\) (1/4)\(^2\) + 20(3/4)\(^3\) (1/4)\(^3\)

= 6(243/1024)(1/4) + 15(81/256)(1/16) + 20(27/64)(1/64)

= 0.36 + 0.30 + 0.13

≈ 0.79

Correct Answer: C. \(\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}\)

Explanation

\(\begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix}\)

= \(\begin{pmatrix} 1\times 0 + 2\times 1 & 1\times 1 + 2\times3 \\ 5\times0 + 1\times1 & 5\times1 + 1\times 3 \end{pmatrix}\)

= \(\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}\)

Correct Answer: D. 2

Explanation

\(\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\log_{5} 8^{\frac{1}{2}}}\)

= \(\frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8}\)

= \(\frac{1}{\frac{1}{2}} \)

= 2

Explanation

If d = deceleration, then retarding force = \(md\).

This is also the force down the plane \(mg \sin \theta\)

\(\therefore md = mg\sin \theta \implies d = g \sin \theta\)

= \(10 \sin 30° = 10 \times \frac{1}{2} = 5 ms^{-2}\)

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

At the highest point, v = 0 ms\(^{-1}\).

\(0^{2} = u^{2} - 2(5)(15) \)

\(u^{2} = 150 \implies u = \sqrt{150} = 12.247 ms^{-1}\)

(ii) \(v = u - dt\)

\(0 = 12.247 - 5t \implies 5t = 12.247\)

\(t = \frac{12.247}{5} = 2.4494 \approxeq 2.45 s\)

(b)(i) Resultant force

\(F = \begin{pmatrix} 10 \cos 0° \\ 10 \sin 0° \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \cos 30° \\ 5\sqrt{3} \sin 30° \end{pmatrix} + \begin{pmatrix} 5 \cos 30° \\ 5 \sin 30° \end{pmatrix} + \begin{pmatrix} 5 \cos 300° \\ 5 \sin 300° \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \cos 330° \\ 5 \sqrt{3} \sin 330° \end{pmatrix}\)

= \(\begin{pmatrix} 10 \times 1 \\ 10 \times 0 \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \times \frac{\sqrt{3}}{2} \\ 5 \sqrt{3} \times \frac{1}{2} \end{pmatrix} + \begin{pmatrix} 5 \times \frac{1}{2} \\ 5 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 5 \times \frac{1}{2} \\ -5 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 5\sqrt{3} \times \frac{\sqrt{3}}{2} \\ -5\sqrt{3} \times \frac{1}{2} \end{pmatrix}\)

= \(\begin{pmatrix} 10 \\ 0 \end{pmatrix} + \begin{pmatrix} 7.5 \\ 2.5 \sqrt{3} \end{pmatrix} + \begin{pmatrix} 2.5 \\ 2.5\sqrt{3} \end{pmatrix} + \begin{pmatrix} 2.5 \\ -2.5 \sqrt{3} \end{pmatrix} + \begin{pmatrix} 7.5 \\ -2.5 \sqrt{3} \end{pmatrix}\)

= \(\begin{pmatrix} 30 \\ 0 \end{pmatrix}\)

Magnitude of resultant = \(\sqrt{30^{2}} = 30N\)

(ii) \(F = ma \implies a = \frac{F}{m}\)

= \(\frac{30}{5} = 6 ms^{-2}\)

Correct Answer: D. \(\frac{2}{3}\)

Explanation

\(\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}\)

= \(\frac{1}{2} + \frac{1}{6} = \frac{2}{3}\).

Hint: Probability of A and B occurring should be greater than probability A or B happening.