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Marina86 [1]
4 years ago
8

Each week, Kathy receives a batch of phones that need repairs. The number of phones that she has left to fix at the end of each

day can be estimated with the equation P = 108 - 23d, where P is the number of phones left and d is the number of days she has worked that week. What is the meaning of the value 108 in this equation?
A.) Kathy starts each week with 108 phones to fix.


B.) Kathy repairs phones at a rate of 108 per hour.


C.) Kathy repairs phones at a rate of 108 per day.
Mathematics
1 answer:
adell [148]4 years ago
6 0
Answer:

A.) Kathy starts each week with 108 phones to fix.
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Answer:

4(7+y) and 4·7+4·y

Step-by-step explanation:

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Nata [24]

Answer:

P = 0.9989

Step-by-step explanation:

In order to do this, I will use the following numbers to make the calculations easier. In this case, We'll say that we have 7 majors and 14 non majors anthopology to present a topic.

This means that in the class we have 21 students.

Now, we choose 5 of them, and we want to know the probability that 1 of them is non major.

First, we need to calculate the number of ways we can select the students in all cases, and then, the probability.

First, we'll use the combination formula, to calculate the number of ways we can select the 5 students out of the 21. We use combination, because it does not matter the order that the students are selected.

C = m! / n!(m - n)!

Where:

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With this expression we will calculate first, how many ways we can choose the 5 students out of 21:

C1 = 21! / 5!(21-5)! = 20,349

Now let's calculate the number of ways you can get the all 5 students are non majors:

C2 = 14! / 5!(14 - 5)! = 2002

Now we need to know the number of ways we can get 4 non majors and 1 major:

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C3' = 14! / 4!(14 - 4)! = 1,001

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Now the way to get 3 non majors and 2 majors, we do the same thing we do to get 4 non majors and 1 major, but changing the numbers. Then the way to get 2 non majors and 3 majors, and finally 1 non major and 4 majors:

3 non majors and 2 majors:

C4 = C4' * C4'' = [14! / 3!(14 - 3)!] * [7! / 2!(7 - 2)!] = 7,644

2 non majors and 3 majors:

C5 = C5' * C5'' = [14! / 2!(14 - 2)!] * [7! / 3!(7 - 3)!] = 3,185

1 non major and 4 majors:

C6 = C6' * C6'' = [14! / 1!(14 - 1)!] * [7! / 4!(7 - 4)!] = 490

Finally to know the probability of getting 1 out of the 5 to be non major, we have to sum all the previous results, and divide them by the ways we can choose the 5 students (C1):

P = 2,002 + 7,007 + 7,644 + 3,185 + 490 / 20,349

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