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Julli [10]
3 years ago
12

Suppose a basketball team had a season of games with the following characteristics: Of all the games, 60% were at-home games. De

note this by H (the remaining were away games). Of all the games, 25% were wins. Denote this by W (the remaining were losses). Of all the games, 20% were at-home wins. Of the at-home games, we are interested in finding what proportion were wins. Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?A. P(H)B. P(W)C. P(H and W)D. P(H | W)E. P(W | H)
Mathematics
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

E. P(W | H)

Step-by-step explanation:

What each of these probabilities mean:

P(H): Probability of the game being at home

P(W): Probability of the game being a win.

P(H and W): Probability of the game being at home and being a win.

P(H|W): Probability of a win being at home.

P(W|H): Probability of winning a home game.

Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?

This is the probability of winning a home game. So the answer is:

E. P(W | H)

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<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

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Step-by-step explanation:hope this helped lol hope u have a wonderful thanksgiving day!!!!!!!

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