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

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There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winnin... show full transcript

Worked Solution & Example Answer:There are five matches on each weekend of a football season - HSC - SSCE Mathematics Extension 1 - Question 6 - 2005 - Paper 1

Step 1

Show that the probability that Megan earns one point for a given weekend is 0.7901, correct to four decimal places.

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Answer

To find the probability that Megan earns one point, we need to calculate the probability that she picks more than half of the 5 winning teams. This means she must pick 3, 4, or 5 winning teams. We can use the binomial distribution formula:

P(X=k) = {n race k} p^k (1-p)^{n-k}

Where:

  • n=5n = 5 (total matches)
  • p = rac{2}{3} (probability of winning)
  • kk is the number of matches won.

We will calculate for k=3k=3, k=4k=4, and k=5k=5:

  1. For k=3k=3:

P(X=3) = {5 race 3} imes igg( rac{2}{3} igg)^{3} imes igg( rac{1}{3} igg)^{2} = 10 imes rac{8}{27} imes rac{1}{9} = rac{80}{243}

  1. For k=4k=4:

P(X=4) = {5 race 4} imes igg( rac{2}{3} igg)^{4} imes igg( rac{1}{3} igg)^{1} = 5 imes rac{16}{81} imes rac{1}{3} = rac{80}{243}

  1. For k=5k=5:

P(X=5) = {5 race 5} imes igg( rac{2}{3} igg)^{5} = 1 imes rac{32}{243} = rac{32}{243}

Now, we add these probabilities together:

P(X>2) = P(X=3) + P(X=4) + P(X=5) = rac{80}{243} + rac{80}{243} + rac{32}{243} = rac{192}{243}

To convert this to four decimal places, we round it to 0.7901.

Step 2

Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places.

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Answer

Now that we have determined the probability that Megan earns a point in one weekend is approximately 0.7901, we can find the probability that she earns one point every week over an eighteen-week season. This can be calculated using:

P(extearningapointeveryweek)=(0.7901)18P( ext{earning a point every week}) = (0.7901)^{18}

Calculating this gives:

ightarrow (0.7901)^{18} ext{ yields approximately } 0.1080$$ Thus, the probability correct to two decimal places is 0.11.

Step 3

Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places.

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Answer

To find the probability of earning at most 16 points during the eighteen-week season, we can use the binomial distribution again.

This means we have to find:

P(Xleq16)=1P(X=17)P(X=18)P(X \\leq 16) = 1 - P(X=17) - P(X=18)

Calculating P(X=17)P(X=17) and P(X=18)P(X=18):

  1. For k=17k=17:

P(X=17) = {18 race 17} imes (0.7901)^{17} imes (0.2099)^{1}

  1. For k=18k=18:

P(X=18) = {18 race 18} imes (0.7901)^{18}

By calculating these probabilities and summing them up, we can find:

P(Xleq16)=1P(X=17)P(X=18)P(X \\leq 16) = 1 - P(X=17) - P(X=18)

Final rounding gives the probability to two decimal places.

Step 4

What is the maximum height the rocket will reach, and when will it reach this height?

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Answer

To find the maximum height of the rocket and the time at which it reaches this height, we can analyze the vertical motion equation:

y=4.9t2+200t+5000y = -4.9t^{2} + 200t + 5000

The maximum height occurs at the vertex of this parabola. We can find the time it takes to reach the maximum height using:

t=b2a=2002(4.9)=2009.820.41 secondst = -\frac{b}{2a} = -\frac{200}{2(-4.9)} = \frac{200}{9.8} \approx 20.41 \text{ seconds}

Substituting this value of t back into the height equation:

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

Calculating this gives us the maximum height.

Step 5

The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?

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Answer

To determine when the pilot can operate the ejection seat, we need to analyze the velocity of the rocket as it descends. The vertical and horizontal components of the rocket's velocity are:

vx=dxdt=200m/sv_x = \frac{dx}{dt} = 200 \, \text{m/s}
vy=dydt=9.8t+200v_y = \frac{dy}{dt} = -9.8t + 200

The angle of descent θ=tan1(vyvx)\theta = \tan^{-1}(\frac{v_y}{v_x})

Setting this between 45° and 60° allows us to find the time intervals where the pilot can eject. We solve for t at these angles.

Step 6

For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than 350 m s$^{-1}$. What is the latest time at which the pilot can eject safely?

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Answer

The speed of the rocket can be calculated using:

v=vx2+vy2v = \sqrt{v_x^2 + v_y^2}

This equation gives the total speed. We set this value less than or equal to 350 m/s to find the latest time at which the pilot can safely eject.

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