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Mathematics

1 answer:

olga55 [171]3 years ago

3 0

A. The formula to find the volume of a cylinder (that's what the first shape is):
pi r^2 h So, when you plug in the numbers and calculate, you get: 24.5pi

The formula to find the volume of a rectangular prism is: length x width x height.
6 x 9 = 54 54 x 2 = 108

V of the first pan = 24.5pi V of the second pan = 108 The second pan has a larger volume.

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A cylinder has a radious of 6 ft and a height of 8 ft what is the volume of the cylinder?

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Answer:

V≈904.78ft³

Step-by-step explanation:

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Which shows a correct order to solve this story problem? Kent and Curtis went to the state fair. They had to pay a total of $7.1

KiRa [710]

The problem is asking how much each person will need to pay. Simplifying the problem into an equation with variables (an algorithm) will greatly help you solve it:

S = Sales Tax = $ 7.18 per any purchase
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F = Food = $ 35.50 purchases for two people

We know the cost for one person was: (22.50) + [(35.50/2) + 7.18] =
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Method A)
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Method A is the correct answer

Method B)
[(2A + (1/2)F + 2S) /2 = [(2)(22.50) + 35.50(1/2) + (2)7.18] / 2 = $38.55
Wrong answer. This method is incorrect because the tax for both tickets bought are not being used in the equation.

Method C)
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3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$