There are five matches on each weekend of a football season - HSC - SSCE Mathematics Extension 1 - Question 6 - 2005 - Paper 1

If the probability that she earns one point in a given week is 0.7901, then the probability that she earns one point every week for 18 weeks is:

$P = (0.7901)^{18}$

Calculating this, we find:

$P \approx 0.090338$

Thus, the probability correct to two decimal places is approximately 0.09.

To find this, we first note that the expected number of weeks she earns points is:

$E[X] = n \times p = 18 \times 0.7901 = 14.1838$

We can use the normal approximation to the binomial distribution:

$P(X \leq 16) \approx P(Z \leq \frac{16 - 14.1838}{\sqrt{18 \times 0.7901 \times (1 - 0.7901)}})$

Calculating:

1. Mean = 14.1838 and variance = (n \times p \times (1 - p) \approx 2.8611)
2. Standard deviation = (\sqrt{2.8611} \approx 1.6922)
3. Z-score = (\frac{16 - 14.1838}{1.6922} \approx 1.07)

Using standard normal distribution tables, we find: $P(Z \leq 1.07) \approx 0.8577$

Thus, the probability of Megan earning at most 16 points is approximately 0.86.

The height of the rocket is given by the equation:

$y = -4.9t^2 + 200t + 5000$

To find the maximum height, we can take the derivative and set it to zero:

$\frac{dy}{dt} = -9.8t + 200 = 0$

Solving for t gives: $t = \frac{200}{9.8} \approx 20.41 \, \text{s}$

Substituting this back to find the height:

$y(20.41) = -4.9(20.41)^2 + 200(20.41) + 5000$

Calculating gives the maximum height as approximately 5500 m.

To find the conditions under which the ejection seat can be operated, we need to find the angle of descent. The angle of descent can be derived from the velocity components:

• Horizontal velocity: $\frac{dx}{dt} = 200$ m/s
• Vertical velocity: $\frac{dy}{dt} = -9.8t + 200$ m/s

The angle of descent is given as:

$\tan(\theta) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Setting the angle condition:

• For 45°: $\frac{dy}{dx} = -1$
• For 60°: $\frac{dy}{dx} = -\sqrt{3}$

The respective conditions give two equations that need to be solved for t.

For the parachute to open safely, the speed of the rocket must be less than or equal to 350 m s$^{-1}$, which can be found by:

1. The total speed of the rocket: $v = \sqrt{(200)^2 + (-9.8t + 200)^2}$
2. Solving the resulting equation $v \leq 350$ for t will provide the latest time at which the pilot can eject safely.