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3 years ago

Write the expression in simplest form using only positive exponents (x^2/3y^-1/2)^-6

1 answer:

5 0

$\left( \cfrac{x^2}{3y^{- \frac{1}{2} }} \right)^{-6}=\left( \cfrac{3y^{- \frac{1}{2} }}{x^2} \right)^{6}=\left( \cfrac{3}{x^2y^{ \frac{1}{2} }} \right)^{6}= \cfrac{3^6}{x^{2*6}y^{ \frac{1}{2}*6 }} = \cfrac{729}{x^{12}y^3}$

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The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago

Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.

andrey2020 [161]

Check the picture below.

notice, the focus point is at 4,5 whilst the directrix line is at y = -3, below the focus point, meaning the parabola is vertical and opening upwards.

keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}})
\\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------$

$\bf \begin{cases}
h=4\\
k=1\\
p=4
\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)
\\\\\\
\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y$

8 0

3 years ago

PLEASE HELP!!!!!!! Which of the following sets of data does not contain an outlier? A.25, 33, 29, 31, 33, 33, 78, 28 B.101, 99,

lesantik [10]

Answer:

Choice C.

Step-by-step explanation:

Choice C has the most numbers close together, and not far away like an outlier.

Please mark me BRAINLIEST! Thanks!

5 0

2 years ago

Read 2 more answers

The length of a rectangle is 10 feet more than the width. The area of the rectangle is 96 square feet. Find the dimensions of th

alexandr1967 [171]

Answer: The dimensions of the rectangle are 6ft and 16ft.

Step-by-step explanation:

Let the width of the rectangle be y

Let the length of the rectangle be y+10

Area= 96ft

Since area of a rectangle is length multiplied by width

y(y+10) = 96

y^2 +10y =96

y^2 +10y -96= 0

Solving algebraically,

y^2 +16y - 6y - 96= 0

y(y+16) - 6(y+16)= 0

(y-6)(y+16)= 0

y-6= 0

y =6ft

Recall that the width was denoted as y. That means the width is 6ft.

Since length is y+10= 6+10 = 16ft

Length is 16ft, width is 6ft.

7 0

2 years ago

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there are 48 students participating in a debateing tournament each team has 3 students how many debating teams are there

babymother [125]

Hello!

Let's divide our 48 students among groups of 3 students below.

48/3=16

Therefore, there are 16 debating teams.

I hope this helps!

3 0

3 years ago

Read 2 more answers