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ICE Princess25 [194]
3 years ago
13

The volume of sphere Q is 50% more than the volume of sphere P. The volume of sphere R is 50% more than the volume of sphere Q.

Find the volume of sphere P as a fraction of sphere R. Write your answer in the form a/b
Mathematics
1 answer:
STALIN [3.7K]3 years ago
6 0
<h2>Use this link to find your answer</h2><h2>http://ldh.la.gov/</h2>
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Assign a number to each student, and use a computer program to generate 100 random numbers between 1 and 2000. Ask those students whose numbers are selected.

Step-by-step explanation:

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In a random sample of 200 cars of a particular model,
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By using proportions, We expect 150 defects out of the 10,000 cars if the 3 defects out of the 200 cars sampled is representative of all 10,000 cars.

Option A is correct.

First, let's set up the proportions.

A proportion is an equation in which two ratios are set equal to each other.

We have 3/200 cars with defects and we want to know X for X/10000.

So 3/200 = X/10000

Now solve for X.

3/200 = X/10000

X = \frac{3}{200} \times 10000

X = 150

So the answer is: We expect 150 defects out of the 10,000 cars if the 3 defects out of the 200 cars sampled is representative of all 10,000 cars.

Option A is correct.

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2 years ago
Potato cause 0.32 per pound how much does 0.75 pounds of sweet potatoes cost
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5 0
3 years ago
Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are
Mariulka [41]

Answer:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(173.6,49.8)  

Where \mu=173.6 and \sigma=49.8

We are interested on this probability

P(141.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0
3 years ago
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