Further Mathematics Questions and Answers

\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)

\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)

\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)

p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')

= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)

= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)

= \(\frac{5}{12}\)

\(v(t) = 3t^{2} - 2t + 1\)

\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)

\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)

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

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

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

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

No explanation has been provided for this answer.

\( P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

\(\begin{vmatrix} 1 \times 1 + 1 \times 2 & 1 \times 1 + 1 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 1 + 1 \times 1 \end{vmatrix}\)

= \(\begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\)

= \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\)