a) 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

We apply the given logistic growth equation:

1. Rearranging, $C = P \cdot \left( 1 + \frac{1}{t}C \right)$

In 1980, assume:

• $P(0) = 150,000$, hence $P = 150,000 e^{r \cdot t}$

At year 20 (2000), we have:

• $P(20) = 600,000$. We set the equations by substituting:

  • Rearranging and solving yields:

  $C \approx 1,130,000$

Let ( y = (x_1, y_1, z_1) ) and ( v = (x_2, y_2, z_2) ).

Using the dot product definition:

1. Calculate ( \boldsymbol{y} \cdot \boldsymbol{v} = x_1 x_2 + y_1 y_2 + z_1 z_2 ).

2. Find ( |\boldsymbol{y}| = \sqrt{x_1^2 + y_1^2 + z_1^2} ) and ( |\boldsymbol{v}| = \sqrt{x_2^2 + y_2^2 + z_2^2} ).

3. Hence, showing the relationship leads to:

   $|\boldsymbol{y}| \cdot |\boldsymbol{v}|^2 = (\text{result from dot product}) \text{ in square form.}$

To find the sample size required for our proportions to meet the acceptable bounds, we use:

1. Define $p$ as ( \frac{3}{500} ).
2. Set up the condition that: $\, \sigma = \sqrt{\frac{p(1 - p)}{n}} \leq\frac{0.025}{\sqrt{n}}$
3. Solving above for n, we get: $n \approx 500 * (z(0.025)^2) \approx 497$

Thus, rounding to the nearest thousand gives us: $n \approx 1000.$

Using the product rule for differentiation:

1. Find $g'(x) = 3x^2 + 4$.
2. Thus, $f'(x) = g'(x) + xg''(x)$. Evaluate at x = 3:
   • Calculate $g''(x)$
   • Combine: $f'(3) = g'(3) + 3g''(3)$
   • Finally evaluate to find the gradient at the specific point.