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disa [49]
3 years ago
6

A waitress works 1.75 hours less in the afternoon than in the evening. If she works 5      1/8

Mathematics
1 answer:
gayaneshka [121]3 years ago
3 0
Let x be the number of hours that the waitress works in the evening.

We know for our problem that she works 5 \frac{1}{8} in the afternoon and that she works <span>1.75 hours less in the afternoon than in the evening, so:
</span>x=5 \frac{1}{8} -1.75
<span>Since </span>\frac{1}{8} =0.125, we can rewrite our expression:
x=5.125-1.75
x=3.375

We can conclude that she works 3.375 hours in the evening, or expressed as a mixed fraction: 3 \frac{3}{8} hours.
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In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
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Answer:

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c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

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Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

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