(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\). (b) The first 5 students to arrive in a school on a Monday ...
Question 1 Report
(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\).
(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :
(i) both were boys ; (ii) the two were of different sexes.
Answer Details
(a) Evaluate \(\dfrac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) with logarithms.
First \(\sqrt{0.00695}\): \(\log 0.00695 = \bar{3}.8420\), half of it \(= \bar{2}.9210\), so \(\sqrt{0.00695} = 0.08337\).
Numerator: \(\log 15.05 = 1.1775\); \(1.1775 + \bar{2}.9210 = 0.0985\), giving \(15.05 \times 0.08337 = 1.2547\).
Divide by \(695\): \(\log 695 = 2.8420\); \(0.0985 - 2.8420 = \bar{3}.2565\).
\[\text{answer} = 1.806 \times 10^{-3} \approx 0.00181\]
(b) Two boys and three girls; two are chosen at random from the five. Total ways \(= \binom{5}{2} = 10\).
(i) Both boys: \(\dfrac{\binom{2}{2}}{\binom{5}{2}} = \dfrac{1}{10}\).
(ii) Different sexes (one boy, one girl): \(\dfrac{2 \times 3}{10} = \dfrac{6}{10} = \dfrac{3}{5}\).
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