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MrRa [10]
3 years ago
5

Over the last three evenings, Yoko received a total of 73 phone calls at the call center. The third evening, she received 3 time

s as many calls as the first evening. The second evening, she received 7 fewer calls than the first evening. How many phone calls did she receive each evening?
Mathematics
1 answer:
stealth61 [152]3 years ago
3 0
The first evening she received x calls.
The second evening she received 7 fewer calls than the first evening, so she received x-7 calls.
The third evening she received 3 times as many calls as the first evening, so she received 3x calls.
Over the three evenings she received 73 calls altogether.

x+x-7+3x=73 \\
5x-7=73 \\
5x=73+7 \\
5x=80 \\
x=\frac{80}{5} \\
x=16 \Leftarrow \hbox{the first evening} \\ \\ \hbox{the second evening:} \\
x-7=16-7=9 \\ \\ \hbox{the third evening:} \\
3x=3 \times 16=48

The first evening she received 16 calls, the second evening she received 9 calls, the third evening she received 48 calls.
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Answer:

a) 5.13% probability that all of the offices are filled by members of the debate​ team

b) 2.56% probability that none of the offices are filled by members of the debate​ team

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the people are chosen is important, since the first person is the president, the second is the vice president and so on... So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

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

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Desired outcomes:

4(president, vice​ president, secretary, and treasurer) offices from a set of 8(candidates of the members debate team).

So

D = P_{(8,4)} = \frac{8!}{(8-4)!} = 1680

Total outcomes:

4 offices from a set of 15 candidates.

T = P_{(15,4)} = \frac{15!}{(15-4)!} = 32760

Probability:

P = \frac{D}{T} = \frac{1680}{32760} = 0.0513

5.13% probability that all of the offices are filled by members of the debate​ team

(b) What is the probability that none of the offices are filled by members of the debate​ team?

Desired outcomes:

4(president, vice​ president, secretary, and treasurer) offices from a set of 7(candidates who are not of the members debate team).

D = P_{(7,4)} = \frac{7!}{(7-4)!} = 840

Total outcomes:

4 offices from a set of 15 candidates.

T = P_{(15,4)} = \frac{15!}{(15-4)!} = 32760

Probability:

P = \frac{D}{T} = \frac{840}{32760} = 0.0256

2.56% probability that none of the offices are filled by members of the debate​ team

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Answer:

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Step-by-step explanation:

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Volume v =

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Radius r =  =\frac{diameter}{2} = \frac{2}{2} = 1 in

Since we are required to solve for the circumference of a circle

The expression for the circumference of a circle is given as

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