A plane needs to travel to a destination that is on a bearing of 063° - HSC - SSCE Mathematics Extension 1 - Question 14 - 2021 - Paper 1

Let $P_0 = 150,000$, then $P = P_0$ when $t = 20$, with $C$ being the carrying capacity. Given that the population in the year 2000 was 600,000:

Substituting into the equation: $P(t)=P_0 e^{(r*t)}$ Substituting values: $600,000 = 150,000 e^{(r*20)}$ Finding $r$: $\implies \ln(4) \sim r*20 \implies r \sim \frac{\ln(4)}{20}$ Now substituting back into the logistic equation, we find: Substituting back leads to: $\frac{C}{150,000} + \frac{1}{600,000} = \frac{1}{C - 600,000}\n$After some algebra, we find $C \approx 1,130,000.$

Let $\mathbf{y} = |\mathbf{v}|^2$ where $\mathbf{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$. Thus, $\mathbf{y} \cdot \mathbf{v} = \begin{pmatrix} x \ y \ z \end{pmatrix} \cdot \begin{pmatrix} x \ y \ z \end{pmatrix} = x^2 + y^2 + z^2 = |\mathbf{v}|^2$

To determine the required sample size $n$ to ensure that $\hat{p}$ does not exceed $0.025$, we utilize the formula: $n = \left( \frac{z_{\alpha/2}^2 \cdot 0.5(1-0.5)}{(0.025)^2} \right)$ Setting $z_{\alpha/2}$ as approximately 1.96. This calculates to: $n = \left( \frac{1.96^2 \cdot 0.5 \cdot 0.5}{0.000625} \right) = 1536$ Rounding gives a sample size of approximately 1600.

First, compute $g'(x)$: $g'(x) = 3x^2 + 4$ Now, evaluate at $x = 3$: $g'(3) = 3(3^2) + 4 = 27 + 4 = 31$ Thus, $f'(x) = g'(x) + xg^{\prime \prime}(x)$ Finding $f'(3)$ gives: $f'(3) = 31 + 3g^{\prime \prime}(3)$ This evaluates to the required gradient.