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Nataly [62]
3 years ago
11

A cylinder and a cone have the same diameter: 8 inches. The height of the cylinder is 3 inches. The height of the cone is 18 inc

hes.
Use π = 3.14.

What is the relationship between the volume of this cylinder and this cone? Explain your answer by determining the volume of each and comparing them. Show all your work.
Mathematics
1 answer:
mina [271]3 years ago
8 0

Vcylinder=\pi r^2 h

Vcone=\frac{1}{3} \pi r^2 h


if we compute the volumes

given that d=8 and height cylinder=3 and height cone=18

note that d/2=r so 8/2=4=r


Vcylidner=[/tex] \pi (4)^2* 3[/tex]=\pi (48)=150.72 cubic inches

Vcone=\frac{1}{3} \pi (4)^2 * 18=\pi (96)=301.44 cubic inches


hmm, if we divide the volume of the cone by thhe volume of the cylinder, we get 301.44/150.72=2


so the relationship is that the cylinder volume is 2 times that of the volume of the cone

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For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les
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Answer:

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

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The margin of error for this case is given by:

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

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

\alpha=0.05 represent the significance level (no given, but is assumed)    

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

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

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

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