Further Mathematics Questions and Answers

\(2\log_{3} 8 - 3\log_{3} 2 = \log_{3} 8^{2} - \log_{3} 2^{3}\)

= \(\log_{3}(\frac{64}{8}) \)

= \(\log_{3} 8 = \log_{3} 2^{3}\)

= \(3 \log_{3} 2\)

\(a . b = |a||b| \cos \theta\)

\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)

\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)

\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)

\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)

\(\theta = \cos^{-1} 0.3162 = 72°\)

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

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

No explanation has been provided for this answer.

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

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

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

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

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

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

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