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Reptile [31]
3 years ago
5

MATH FRACTIONS!! What is 3/4 divided by 5/6 how to solve it the step by step. I NEED IT

Mathematics
2 answers:
Reil [10]3 years ago
6 0

Answer:

9/10

Step-by-step explanation:

the rule says

\frac{a}{b} /\frac{c}{d} =\frac{ad}{bc}

so in our case

a=3

b=4

c=5

d=6

so the division is

\frac{3*6}{4*5} \\\\=\frac{18}{20}

we can simplify

\frac{18}{20} \\\\=\frac{2(9)}{2(10)} \\\\=\frac{9}{10}

that's our answer

I don't know if you need step by step of the rule, if you need let me know

vaieri [72.5K]3 years ago
4 0

Answer:

9/10

Step-by-step explanation:

hi again haha so basically apply what i already told you

flip 5/6 to 6/5 and change division to x

so 3/4 x 6/5 both 4 and 6 can be divided by 2 so ask how many times does 2 go into 4 and 6 so it simplifies to 3/2 x 3/5

multiply the both tops and bottoms to get 9/10 :)

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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
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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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Step-by-step explanation:

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