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Licemer1 [7]
3 years ago
9

So there's this cone stuff and i needa know

Mathematics
1 answer:
Bas_tet [7]3 years ago
7 0

Hey there! I'm happy to help!

To find the volume of a cone, you take the area of the base, multiply it by the height and divide it by three.

We see that the radius of our base is three. To find the area of the base, we will square the radius and multiply it by 3.14 (formula for area of circle).

3²=9

9×3.14= 28.26

We multiply it by the height.

28.26(2)=56.52

We divide it by three.

56.52/3=18.84

Therefore, the area is 18.84 units cubed.

Now you can find the volume of cones! I hope that this helps! Have a wonderful day!

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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.
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Answer:

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.1 = 0.90, so z = 1.28

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28*\frac{2.3}{\sqrt{717}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 18.5 - 0.1 = 18.4 gallons per day.

The upper end of the interval is the sample mean added to M. So it is 18.5 + 0.1 = 18.6 gallons per day.

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

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3 years ago
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First off, let's convert the decimal to a fraction, notice, we have two decimals, so we'll use in the denominator, a 1 with two zeros then, two decimals, two zeros, thus   \bf 1.\underline{75}\implies \cfrac{175}{1\underline{00}}\implies \cfrac{7}{4}\implies \stackrel{ratio}{7:4}

now, we know then the ratio dimensions for the new photograph, 

\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\
\cfrac{7}{4}\implies \cfrac{4+3}{4}\implies \cfrac{4}{4}+\cfrac{3}{4}\implies 1+\boxed{\cfrac{3}{4}}\impliedby \textit{perimeter is }\frac{3}{4}\textit{ larger}
\\\\\\
\stackrel{areas'~ratio}{\cfrac{s^2}{s^2}}\implies \cfrac{3^2}{4^2}\implies \cfrac{9}{16}\impliedby \textit{area is }\frac{9}{16}\textit{ larger than original}
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3 years ago
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