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hoa [83]
3 years ago
9

A cylindrical piece of iron pipe is shown below. The wall of the pipe is 0.75 inch thick, and the pipe is open at both ends: The

figure shows a cylinder of height 12 inches and diameter 8 inches. What is the approximate inside volume of the pipe? (8 points)
398 cubic inches
603 cubic inches
565 cubic inches
412 cubic inches
Mathematics
1 answer:
const2013 [10]3 years ago
3 0

Answer:

It's 398

Step-by-step explanation.

Diameter of 8.

height of 12.

8 - .75 - .75= 6.5

6.5/2 = 3.25

3.25 x 3.25 x 12 x 3.14 = 397.995

round to the nearest whole number and you get

398

Take the diameter (8) an subtract is by both sides of the walls (0.75 each wall) that gives you 6.5. Take that 6.5 and half that, because you need to have the radius to solve volume problems, giving you 3.25.  Then multiply 3.25 by 3.25 together ( I know it seems redundant but it just works easier this way) This gives you 10.5625. After that take 10.5625 then multiply it into 12, being the height, which gives you 126.75. Finally take 126.75 and multiply that into 3.14 (pie) giving you the answer.

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Answer:

x= (y-5)/2

Step-by-step explanation:

Isolate the x variable.

y= 2x +5

Subtract 5 from both sides.

y-5=2x

Divide both sides by 2.

(y-5)/2= x

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
slega [8]
 <span>Let r(x,y) = (x, y, 9 - x^2 - y^2) 

So, dr/dx x dr/dy = (2x, 2y, 1) 

So, integral(S) F * dS 
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6 0
3 years ago
The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a sta
nordsb [41]

Answer:

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

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 normally distributed samples are 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.

In this problem, we have that:

\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

This is the pvalue of Z when X = 8.2 + 0.02 = 8.22 subtracted by the pvalue of Z when X = 8.2 - 0.02 = 8.18. So

X = 8.22

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

By the Central Limit Theorem

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

Z = \frac{8.22 - 8.2}{0.018}

Z = 1.11

Z = 1.11 has a pvalue of 0.8665

X = 8.18

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

Z = \frac{8.18 - 8.2}{0.018}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

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