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Alexandra [31]
2 years ago
7

Please Show Your Work

Mathematics
1 answer:
Morgarella [4.7K]2 years ago
7 0

The area of the trapezoid will be 32.5 square feet. Then the volume of the geometry will be 341.25 cubic feet.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

The figure is shown.

Then the volume of the given geometry will be

Volume = Area of the trapezoid × Height

Then the area of the trapezoid will be

\rm Trapezoid \ area =  \dfrac{1}{2} \times (8 + 5) \times 5\\\\\\Trapezoid \ area =  32.5 \ ft^2

Then the volume of the geometry will be

Volume = 32.5  × 10.5

Volume = 341.25 cubic ft.

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

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The cost of making a cheese cake is RM45. In
mars1129 [50]

Answer:

CHEESE

Step-by-step explanation:

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3 0
3 years ago
Lin earned $50 last week. Andre earned 80% of what Lin earned. How much did Andre earn last week? *
DaniilM [7]

Answer:

$40

Step-by-step explanation:

Turn the percentage into a decimal.

80% = 0.8

Multiply.

50 * 0.8 = 40

Best of Luck!

5 0
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

3 0
3 years ago
Kenzie has. 4 pages of homework to do. If she can finish 1/3 of a page in one hour how many hours will her homework take?
Maru [420]

Answer:

It will take her 4 1/3 hours

Step-by-step explanation: so you know you needed to  add 4 and 1/3 and you turn the 4 into 4/1 and 1/3 and since it adding you have to have the same denominator and you multiply 1x3 and if you do it on the bottom you have to do it on top so you also do 4x3 and 4x3 is 12 and one 1x3 is 3 and so you don't have to have to do it on the other fraction because now they have the same denominator now and now you add 12/3 + 1/3 and you shall get 13/3 and then you divide 13 divided by 3 and you shall get 4 1/3 and that shall be your answer.

hope it helps you!

5 0
3 years ago
Evaluate the following double integral where a = 2y
Keith_Richards [23]

Change the order of integration.

\displaystyle \int_0^1 \int_{2y}^2 \cos(x^2) \, dx \, dy = \int_0^2 \int_0^{x/2} \cos(x^2) \, dy \, dx \\\\ ~~~~~~~~ = \int_0^2 \cos(x^2) y \bigg|_{y=0}^{y=x/2} \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^2 x \cos(x^2) \, dx

Substitute u=x^2 and du=2x\,dx.

\displaystyle \frac12 \int_0^2 x \cos(x^2) \, dx = \frac14 \int_0^4 \cos(u) \, du = \frac14 \sin(u) \bigg|_{u=0}^{u=4} = \boxed{\frac{\sin(4)}4}

4 0
1 year ago
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