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3 years ago

The volume of a cylinder is 4πx3 cubic units and its height is x units. Which expression represents the radius of the cylinder,

in units?

Mathematics

1 answer:

olga55 [171]3 years ago

4 0

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad 
\begin{cases}
r=radius\\
h=height\\
-----\\
V=4\pi x^3\\
h=x
\end{cases}\implies 4\pi x^3=\pi \cdot r^2\cdot x
\\\\\\
\cfrac{4\pi x^3}{\pi x}=r^2\implies \implies 2^2x^2=r^2\implies \sqrt{2^2x^2}=r\implies 2x=r$

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Approximately 55% of enrolled students are women.

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To find the percentage of a given set of data, you need to divide the number of actual students by the total number of students. In this case, they are wanting the percentage of just women students at the university:

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

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A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

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