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Yuri [45]
4 years ago
14

Scientists are drilling a hole in the ocean floor to learn more about the​ Earth's history. ​ Currently, the hole is in the shap

e of a cylinder whose volume is approximately 4300 cubic feet and whose height is 4.7 miles. Find the radius of the hole to the nearest hundredth of a foot. ​ (Hint: Make sure the same units of measurement are​ used.)
Mathematics
2 answers:
blagie [28]4 years ago
8 0

volume of cylinder=πr²h

4,300=22/7×r×r×4.7

4,300=3.14×4.7×r²

4300=14.758×r²

291.36=r²

17.069 =r

Misha Larkins [42]4 years ago
5 0

Answer:

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

π is a constant whose value is 3.14

r represents the radius of the cylinder.

h represents the height of the cylinder.

From the information given, the volume of the cylindrical hole is

approximately 4300 cubic feet and its height is 4.7 miles. Converting 4.7 miles to feet,

1 mile = 5280 feet

4.7 miles = 4.7 × 5280 = 24816 feet

Therefore

4300 = 3.14 × r² × 24816 = 77922.24r² = 4300

r² = 4300/77922.24 = 0.055

r = √0.055 = 0.23 feet

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Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

Z = \frac{X - \mu}{s}

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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