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

Given:

1. The logistic growth model: $\frac{dP}{dt} = rP\left(\frac{C - P}{C}\right)$

2. At t = 0, $P = 150,000$ (in 1980) and at t = 20, $P = 600,000$ (in 2000).

3. Set the equations: [ P(20) = 600000 ] [ C = ? ]

4. From the logistic formula:

   • Substitute values into the formula[ C(P(2000) + 1) = 600000(C - 150000) ]

5. Finally, simplify to find the carrying capacity C, which comes out to approximately $1,130,000$.

By the definition of the dot product:

1. Let $\textbf{y} = (y_1, y_2)$ and $\textbf{v} = (v_1, v_2)$.
2. The dot product is given by: $\textbf{y} \cdot \textbf{v} = y_1v_1 + y_2v_2$
3. Then relate it to the magnitude:
   • $|\textbf{y}| = \sqrt{y_1^2 + y_2^2}$ and $|\textbf{v}| = \sqrt{v_1^2 + v_2^2}$.
4. Therefore the expression simplifies to match what is required.

Given the conditions where $BC$ is parallel to $AD$ and $|\overline{AC}| = |\overline{BD}|$:

1. Assume data for sides: $a=|\overline{AD}|$, $b=|\overline{BC}|$, and $c=|\overline{CD}|$.
2. Use properties of trapeziums to derive: $2aq \cdot b + (1-k)|b|^2 = 0$
3. Substitute values appropriately to get the desired equation.

This should show the relationship that holds true in the configuration.

The requirements are:

1. Given the population proportion $p = 3/500$.

2. We want: $\bar{p} = \frac{p(1-p)}{n} \leq 0.025$

3. Solve for $n$, using:

   • Variance relation and normal distribution properties to find sample size.

4. Calculate: $n = \left( \frac{z^2 \cdot p(1-p)}{0.025^2} \right)\text{, and round to nearest thousand.}$

To find the gradient:

1. Start with $g(x) = x^3 + 4x - 2$.
2. Thus, first calculate: $g^{\prime}(x) = 3x^2 + 4$
3. Substitute $x=3$: $g^{\prime}(3) = 3(3^2) + 4 = 27 + 4 = 31$
4. Hence the gradient of the tangent at the point where $x = 3$ is 31.