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Orlov [11]
3 years ago
12

Will and Thomas are collecting baseball cards. Will already has ten baseball cards and plans to collect 5 cards per month. Thoma

s has twenty baseball cards and plans to collect 3 cards per month. After how many months will Thomas and Will have the same number of baseball cards?
Mathematics
2 answers:
jonny [76]3 years ago
8 0

Answer: 5 months

Step-by-step explanation:

okay we set up and equation.

10 + 5m is how many cards will gets, where m is months

20 + 3m is how many cards thomas gets, where m is months.

lets see how many months it will take for them to be equal:

10 + 5m = 20+ 3m

5m = 10 +3m

2m = 10

m = 5

it will take 5 months for them to have the same amount of cards.

andre [41]3 years ago
5 0
I got 5 months, because you add on to each person frequency until you hit the same amount (sorry if it’s a sad answer)
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The probability that Barry Bonds hits a home run on any given at-bat is 0.16, and each at-bat is independent.
serious [3.7K]

Answer:

a) 0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

b) The expected number of at-bats until the next home run is 6.25.

Step-by-step explanation:

For each at bat, there are two possible outcomes. Either it is a home run, or it is not. The probability of an at bat resulting in a home run is independent of any other at-bat, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that Barry Bonds hits a home run on any given at-bat is 0.16

This means that p = 0.16

Part A: What is the probability that the next home run will be on his fifth at-bat?

0 on his next 4(P(X = 0) when n = 4)

Home run on his 5th at-bat, with 0.16 probability. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{4,0}.(0.16)^{0}.(0.84)^{4} = 0.49787136


0.49787136 *0.16 = 0.0797.

0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

Part B: What is the expected number of at-bats until the next home run?

The expected number of trials for n successes is given by:

E = \frac{n}{p}

In this question, n = 1, p = 0.16. So

E = \frac{1}{0.16} = 6.25

The expected number of at-bats until the next home run is 6.25.

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Step-by-step explanation:

Find the attachments for complete step by step explanation

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The answer would be 400 meters in width.

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