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NeTakaya
3 years ago
9

you send out 20,000 emails of 6% are opened of those 9% clicked on the link to register for something of those who clicked on th

e link 30% complete the registration how many people completed the registration​
Mathematics
2 answers:
nikitadnepr [17]3 years ago
6 0

Answer:

<u>Rounding to the next whole, 33 persons completed the registration</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of e-mails sent out = 20,000

Number of e-mails opened = 6%

Number of e-mails that clicked on the link to register = 9% of the 6%

Number of e-mails that completed the registration = 30% of 9% of 6%

2.  How many people completed the registration​?

Let's do all the calculations to answer the question, this way:

Number of e-mails opened = 6 % of 20,000

Number of e-mails opened =0.06 * 20,000 = 1,200

Number of e-mails that clicked on the link to register = 9% of the 6%

Number of e-mails that clicked on the link to register = 0.09 * 1,200 = 108

Number of e-mails that completed the registration = 30% of 9% of 6%

Number of e-mails that completed the registration = 0.3 * 108 = 32.4

<u>Rounding to the next whole, 33 persons completed the registration</u>

maxonik [38]3 years ago
4 0

Answer:

The answer is 32.

Step-by-step explanation

hope this helps :D

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gavmur [86]

Answer:

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

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

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:

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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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