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andre [41]
2 years ago
13

The question is, "Find the volume of the bodies given in the figures below, which have a hole along their entire length." How do

I measure it the volume in this case?
As you can see, height is 9cm, length is 12, and width is 18 . ​

Mathematics
1 answer:
aniked [119]2 years ago
4 0

Answer:

1782 cm³

Step-by-step explanation:

First, you measure the volume of the cuboid as a whole which is length x breadth x height, 9 x 12 x 18 = 1944 cm³. Then measure the volume occupied by the space, 3 x 3 x 18 since the hole runs along the entire length = 162 cm³.

Then you subtract the volume occupied by the hole from the volume of the cuboid in total, 1944 - 162 = 1782 cm³

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Simplify completely 3x+18/18 ...?
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Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc
Dima020 [189]

Answer:

a) ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

b) ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=\pm 1.96

If we replace the values into equation (a) for 1983 we got:

ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

Part b

Since is the same confidence level the z value it's the same.  

If we replace the values into equation (a) for 2008 we got:

ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

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I hope this is right.....

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Have an awesome day :)
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