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Aleks [24]
3 years ago
10

A chocolate bar measures 30 mm wide, 70 mm long, and (or 4.5) mm high. Will's Wrapping Company is making a wrapper to cover the

chocolate bar. Use the correct net and determine how much paper will be needed to make the wrapper.
Mathematics
1 answer:
KiRa [710]3 years ago
8 0

Hey there! I'm happy to help!

First, let's find the area of the top of the bar.

30×70=2100

This is the same as the bottom of the bar, so we have another 2100 mm, giving us 4200 in total.

Now we want to find the two long sides. This is 70×4.5, which is 315. Since there are two of these sides, we have another 315, giving us 630 in total there.

Now we have the short sides.

30×4.5=135

135×2=270

Now, we add up all of these measurements.

4200+630+270=5100

Therefore, Will's Wrapping Company will need 5100 mm² of paper to make their wrapper.

Have a wonderful day! :D

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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

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Answer:

5.5 inches

Step-by-step explanation: hope it helps

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