(i) Conditions on $V$: $V$ is a constant.
(iii) $I = \frac{3A}{2\mathrm{e}}$.
(i)(a) $102.43.
(i)(b) $1215.71.
(i)(c) First day of June 2018.
(ii)(a) $$(100+12b)$.
(ii)(b) $b=\frac{4}{3}$.
(iii) $b=1.23$.
Paper 2
(i) $f(x)=\frac{2\sqrt{2}}{9}(x+18)^{\frac{3}{2}} + 45$.
(ii) $(54,237)$.
(a)(i) $x=2+3i, 2-3i$ or $\frac{1}{2}$.
(b)(i) $\frac{-3+3\sqrt{3}i}{2}$ and $\frac{-3-3\sqrt{3}i}{2}$.
(b)(ii) $3\mathrm{e}^{0i}, 3\mathrm{e}^{\frac{2\pi}{3}i},3\mathrm{e}^{\frac{-2\pi}{3}i}.$
(b)(iii) Sum = $0$, product = $27$.
(i) $D(-5,-4,3)$.
(ii) $4x+45y+20z=200$.
(iii) $58.6^{\circ}.$ (iv) $6.88$ units.
(i) $-2x^2-\frac{4}{3}x^4-\frac{64}{45}x^6$.
Expansion not valid when $x=\frac{\pi}{4}$.
(ii) $-2x-\frac{4}{9}x^3-\frac{64}{225}x^5+c$.
$-1.0644$.
(iii) $-1.0670$.
(i) The sample size should be at least 30 so that $n$ is large and Central Limit Theorem will apply. Hence $\overline{X}$ will be normally distributed approximately. The fans should be randomly chosen.
(ii) Let $X$ be time to failure and $\mu$ the population mean of $X$.
Null hypothesis $H_0: \mu = 65000$ Alternative hypothesis $H_1: \mu < 65000$.
(iii) $\sigma^2 > 9420000$.
(i) Hint: use a binomial.
(ii) $\frac{5}{9} < p < \frac{2}{3}$.
(iii) $0.430$.
(i) $P(A’ \cap B’) = 1-a-b+ab$. Since $P(A’) \cdot P(B’) = P(A’ \cap B’)$, $A’$ and $B’$ are independent.
(ii) $P(A’ \cap B’) = 1-a-c$.
(iii) $\frac{2}{15} \leq P(A \cap B) \leq \frac{1}{3}$.
(i) $P(S=6) = \frac{2}{(n+4)(n+5)},P(S=7) = \frac{12}{(n+4)(n+5)}$, $P(S=8) = \frac{4n+6}{(n+4)(n+5)},P(S=9) = \frac{6n}{(n+4)(n+5)}$, $P(S=10) = \frac{n^2-n}{(n+4)(n+5)}$.
(ii) $P(S=10)=0$.
This is because it is impossible to obtain $S=10$ when there is only one ball numbered 5 (at least two balls numbered 5 are needed to obtain $S=10$.
(iii) $Var(S)=\frac{22n^2+78n+36}{(n+4)(n+5)^2}$.
(i) It is unlikely to be modelled by $P=aR+b$ because the scatter diagram shows that the rate of increase of $P$ with respect to $R$ is increasing as $R$ increases.
(ii) For $P$ and $R$, $r=0.969$.
For $P$ and $R^2$, $r=0.993$. Since the $|r|$ value for $P$ and $R^2$ is closer to 1, $P=aR^2+b$ is the better model.
$P=2.85\times 10^{-8}R^2-0.283$.
(iii)$R=6450$. The estimate is reliable since $P=0.9$ is within the given data range and $|r|\approx 1$.
(iv) $P=0.0273$. The estimate is not reliable since $R=3300$ is outside the given data range.
(v) $P = 1.02 \times 10^{-4} R^2 – 0.283$.
(ii) $0.605$.
(iii) $0.773$.
(iv) $0.126$.
(v) $136$.
(vi) $0.185$.