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Leya [2.2K]
3 years ago
13

What is 12/25 as a percent

Mathematics
2 answers:
mart [117]3 years ago
7 0
12/25 is 48% becuase 12/25 is alao 48/100 because 12 x 4= 48 and 25 x 4= 100.
Sonbull [250]3 years ago
5 0
To make a fraction into a percent you have to divide the numerator by the denominator and multiply by 100. So 12/25 is .48 then when multiplied by 100 you get: 48%  Hope it helps!!
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Leo bought 6 chew toys for his new puppy each chew toy cost $4 how much did Leo spend for the chew toys
Nikolay [14]
6*4 equals 24 bam easy mat
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3 years ago
Read 2 more answers
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

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3 years ago
Hey plz anyone inbox me ​
Scilla [17]

Answer:

Hi

Step-by-step explanation:

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mariarad [96]

Answer:

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Step-by-step explanation:

90°-40°=50°

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Hope this helps! :D

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A bag contains 6 red marbles, 4 black marbles, and 5 blue
mariarad [96]

Answer:

Likely

Step-by-step explanation:

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