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yaroslaw [1]
2 years ago
10

A large diamond with a mass of 712.04 grams was recently discovered in a mine. if the density of the diamond is 3.51 grams over

centimeters cubed, what is the volume? round your answer to the nearest hundredth. (5 points)
question 8 options:

1)

202.9 cm3

2)

507.02 cm3

3)

717.06 cm3

4)

3574.44 cm3
Mathematics
1 answer:
fredd [130]2 years ago
8 0

Answer:

202.9 cm^3

Step-by-step explanation:

Mass is 712.04 grams

Density is 3.51 grams/cm^3

Volume = (712.04 g)/(3.51 grams/cm^3) = 202.9 cm^3

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Esessment
Klio2033 [76]

The area of the given composite figure is 34 square feet

<h3>Area of rectangle</h3>

Area of the rectangle = length * width

<h3>Get the area of the given composite figure</h3>

For the composite figure shown:

Area = (8*3) + (2 * 5)

Area = 24 + 10

Area = 34 square feet

Hence the area of the given composite figure is 34 square feet

Learn more on area of rectangle here: brainly.com/question/25292087

3 0
2 years ago
What is the simplest form for 15:25​
vazorg [7]

Answer:

3/5

Step-by-step explanation:

Both can be divided by 5. So, 15 divided by 5 is 3, and 25 divided by 5 is 5! Hoped this helped :)

3 0
2 years ago
1.A house worth $250,000 has an coinsurance clause of 90 percent. The owners insure the property for $191,250. They then have a
Arisa [49]

1. Multiply the value of the house by the percent:

250,000 x 0.9 = 225,000

Divide the amount the owner insured by the clause amount:

191,250/225,000 = 0.85

Multiply that by the amount of damage:

0.85 x 80,000 = 68,000

They will receive $68,000

2. Follow the same steps as above:

180,000 x 0.75 = 135,000

101,250 / 135,000 = 0.75

0.75 x 50,000 = 37,500

They will receive $37,500

8 0
3 years ago
Read 2 more answers
A Confidence interval is desired for the true average stray-load loss mu (watts) for a certain type of induction motor when the
Vaselesa [24]

Answer:

A sample size of 35 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the width W as such

W = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large must the sample size be if the width of the 95% interval for mu is to be 1.0:

We need to find n for which W = 1.

We have that \sigma^{2} = 9, then \sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3. So

W = z*\frac{\sigma}{\sqrt{n}}

1 = 1.96*\frac{3}{\sqrt{n}}

\sqrt{n} = 1.96*3

(\sqrt{n})^2 = (1.96*3)^{2}

n = 34.57

Rounding up

A sample size of 35 is needed.

3 0
3 years ago
2 + 0.25x what is the answer to this question
zimovet [89]

Answer:

the total late fee

Step-by-step explanation:

the x will stand for each day and 0.25 is the cost so you will multiply 0.25 by x to get the total overdue

3 0
3 years ago
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