\(1\)
\(\frac{1}{2}\)
\(\frac{5}{12}\)
\(\frac{1}{12}\)
Correct answer is C
\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)
\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)
\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)
p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')
= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)
= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)
= \(\frac{5}{12}\)
\(26 ms^{-2}\)
\(18 ms^{-2}\)
\(17 ms^{-2}\)
\(16 ms^{-2}\)
Correct answer is D
\(v(t) = 3t^{2} - 2t + 1\)
\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)
\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)
13 metres
15 metres
18 metres
23 metres
Correct answer is A
\(v^{2} = u^{2} + 2as\)
\(v^{2} = u^{2} - 2gs\)
\(0 = 10^{2} - 2(10)s \implies -100 = -20s\)
\(s = 5 m + 8 m = 13m\) (8m is the height from where it was thrown)
Which of the following is the semi- interquartile range of a distribution?
\(Mode - Median\)
\(\text{Highest score - Lowest score}\)
\(\frac{1}{2}(\text{Upper quartile - Median})\)
\(\frac{1}{2}(\text{Upper quartile - Lower quartile})\)
Correct answer is D
No explanation has been provided for this answer.
If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
\(\begin{vmatrix} 4 & 3 \\ 6 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)
\(\begin{vmatrix} 2 & 2 \\ 6 & 2 \end{vmatrix}\)
\(\begin{vmatrix} 3 & 2 \\ 6 & 4 \end{vmatrix}\)
Correct answer is B
\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)
= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)