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zysi [14]
3 years ago
13

Just help tbh I’m tired

Mathematics
2 answers:
dmitriy555 [2]3 years ago
3 0
She put the decimal point in the wrong soy
Doss [256]3 years ago
3 0

Answer:

Misplaced decimal position

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Answer:

To answer this question, there needs to be choices provided since the description you provided states "which of the following".

Step-by-step explanation:

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What is 2.33 written as a mixed number in simplest form?
aleksandr82 [10.1K]
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hallie is trying to win the grand prize on a game show. Should she try her luck by spinning a wheel with 6 equal sections labele
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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
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
3 years ago
HELP FOR 10 POINTS!!!
Kay [80]
20×8=160
5×8=40
10×8=80
4×8=32
15×8=120

160+40+80+120+32=432
432×2=864
864 is the answer,so your answer is b
6 0
3 years ago
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