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1 year ago

Find the surface area of a hemisphere that has a volume of 486π

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Mathematics

1 answer:

Xelga [282]1 year ago

6 0

$\textit{volume of a hemisphere}\\ V=\cfrac{4\pi r^3}{3}\cdot \cfrac{1}{2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=486\pi \end{cases}\implies 486\pi =\cfrac{4\pi r^3}{6} \implies 6(486\pi )=4\pi r^3 \\\\\\ \cfrac{6(486\pi )}{4\pi }=r^3\implies 729=r^3\implies \sqrt[3]{729}=r\implies \boxed{9=r} \\\\[-0.35em] ~\dotfill$

$\textit{surface area of a hemisphere}\\\\ SA=4\pi r^2\cdot \cfrac{1}{2}\implies \stackrel{\textit{we know that r = 9}}{SA=2\pi (9)^2}\implies SA=162\pi \implies SA\approx 508.94$

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2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

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2 years ago

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xenn [34]

Answer:

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Step-by-step explanation:

3x=30

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2 years ago

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vlada-n [284]

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2 years ago

The radius of one sphere is twice as great as the radius of a second sphere. Find the ratio of their volumes.

Juliette [100K]

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So if the radius of the first sphere is 2r, then it would be (2r)^2 = 4r^2 

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or 4.

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2 years ago