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storchak [24]
3 years ago
13

Find the Sum of (n+8)+(n-12)

Mathematics
1 answer:
Alexxandr [17]3 years ago
4 0

Answer: 2n - 4 is the simplest answer

Step-by-step explanation:

(n + 8) + (n - 12)

8 - 12 = -4

n + n = 2n

2n - 4

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Answer:

c. 15in

Step-by-step explanation:

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3 years ago
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How is 21\30 bigger then 2\3
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By reducing 21/30 you dived 21 and 30 by 3 and you will get 7/10. Is 7/10 larger than 2/3? If we put these fractions in decimal form we will see 7/10 turns into 0.7 and 2/3 turns into 0.667. So the answer is 21/30 is larger than 2/3.
3 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
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:

                                                     Purchased Gum      Kept the Money   Total

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

3 0
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Julie works for 48 hours per week for 12 weeks during the summer, making $5000. If she works for 48 weeks during the school year
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She would need to work 12 hours per week. 
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What is the probability that in a class of 20 students, 2 of them will have the same birthday?
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There would be a 10% chance that they would have the same birthday
8 0
4 years ago
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