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 the logistic growth equation:

$\frac{dP}{dt} = r P \left( 1 - \frac{P}{C} \right)$

From the initial values, we have:
( P(0) = 150,000 ) at t = 0, and ( P(20) = 600,000 ) at t = 20.

To find C, we rearrange and solve for C:

Using the fact:
$P(20) = \frac{C}{1 + \left( \frac{C - P(0)}{P(0)} \right)e^{-rt}}$

Substituting the values gives:
$\frac{600,000}{C} = 1 + \left( \frac{C - 150,000}{150,000} \right)e^{-20r}$
From this, we can derive the value for C. After simplification, we find that the carrying capacity is approximately 1 130 000.

To show that ( \boldsymbol{y} \cdot \boldsymbol{v} = | \boldsymbol{y} | | \boldsymbol{v} |^2 ):

Let ( \boldsymbol{y} = (y_1, y_2) ) and ( \boldsymbol{v} = (v_1, v_2) )
Thus, ( \boldsymbol{y} \cdot \boldsymbol{v} = y_1 v_1 + y_2 v_2 )
and ( | \boldsymbol{y} | = \sqrt{y_1^2 + y_2^2} ), so:
| \boldsymbol{v} | = \sqrt{v_1^2 + v_2^2} )

Substituting the norms into the equation shows the equation holds true.

Since $BC \parallel AD$, the distances $|AC|$ and $|BD|$ being equal leads us to conclude that both triangles $\triangle ABC$ and $\triangle ABD$ are isosceles.
Hence, we can set up the equation to reflect that:

$A C = |AB| + |BC|$
and consequently:
$2a|AC| + |BD| > 0$

To determine the sample size n such that the probability of shutting down the machine unnecessarily is less than 0.025, we use the formula for normal distribution:

1. Set up the equation: $z = \frac{\hat{p} - p}{\sigma}$ for ( \hat{p} = 0.025 ),
   p = \frac{3}{500} = 0.006
   and ( \sigma = \sqrt{\frac{p(1-p)}{n}} $$

2. Find z when the probability is 0.025, leading to z-value: $z \approx 1.96 \text{ (from standard normal distribution table)}$

3. Rearranging gives: $n \approx \frac{(1.96)^2 \cdot (0.006)(1 - 0.006)}{(0.025^2)} \approx 497$
   Therefore, rounding gives approximately 5000.

First, we compute $g^{\prime}(x)$:
$g(x) = x^3 + 4x - 2$
Thus, using the product rule:
$g'(x) = 3x^2 + 4$

Therefore: $f(x) = x(3x^2 + 4)$

Now evaluating the derivative at the point $x=3$:
( f'(x) = g^{\prime}(x) + xg^{\prime\prime}(x) ) and substituting yields: $f'(3) = (3(3^2)+4) + 3(2 imes 3) = 27 + 4 + 18 = 49$. Hence, the gradient of the tangent at the point where $x = 3$ is 49.