The random variables X and Y are jointly continuous, with a joint PDF of the form fX,Y(x,y)={cxy,0,if 0≤x≤y≤1,otherwise, where c is a normalizing constant. For x∈[0,0.5] , the conditional PDF fX|Y(x|0.5) is of the form axb . Find a and b .

Answers

Answer 1

Answer:

The value of a and b are 8 and 1 respectively

Step-by-step explanation:

[tex]f_{x,y}(x,y)=\left\{\begin{matrix}cxy ,\text{if} 0 \leq x \leq y \leq 1\\ 0 , \text{otherwise}\end{matrix}\right.[/tex]

[tex]\int_{0}^{1}\int_{x}^{1}cxy \text{dy dx}=1\\\\c\int_{0}^{1}x\left ( \frac{y^2}{2}\right )_{x}^{1} \text{ dx}=1\\\\\frac{c}{2}\int_{0}^{1}x\left ( 1-y^2\right )\text{dx}=1\\\\\frac{c}{2}\int_{0}^{1}(x-x^3)\text{dx}=1\\\\\frac{c}{2}\left [ \frac{x^2}{2}-\frac{x^5}{4} \right ]_{0}^1=1\\\\\frac{c}{2}\left [ \frac{1}{2}-\frac{1}{4} \right ]=1\\\\\frac{c}{2}\left ( \frac{1}{4} \right )=1\\\\c=8[/tex]

Conditional pdf of [tex]f_{x,y}(x|0.5)[/tex]

[tex]f_{x,y}(x|0.5)=\frac{f_{xy}(x,y=0.5)}{f_{y}y}[/tex]

Normalizing pdf of 1

[tex]f_{y}y=\int_{0}^{y} 8yx \text{dx}[/tex]

[tex]f_{y}y=8y[\frac{x^2}{2}]_{0}^{y}\\f_{y}y=4y(y^2)\\f_{y}y=4y^3\\f_{x|y}(x|0.5)=\frac{8x(0.5)}{4(0.5)^3}=\frac{2x}{0.25}=8x[/tex]

We are given that PDF [tex]f_{x|y}(x|0.5)[/tex] is of the form [tex]ax^b[/tex]

So, on comparing 8x with [tex]ax^b[/tex]

So, a = 8 , b = 1

So, the value of a and b are 8 and 1 respectively

Answer 2

The resulting value of a and b given  the conditional PDF of fX|Y(x|0.5) is mathematically given as

a = 8 , b = 1 .

What are the values of a and b?

Question Parameters:

a joint PDF of the form fX,Y(x,y)={cxy,0,if 0≤x≤y≤1,

For x∈[0,0.5]

the conditional PDF fX|Y(x|0.5)

Generally, the equation for the function of (x,y)   is mathematically given as

[tex]f_{x,y}(x,y)=\left\{\begin{matrix}cxy ,\text{if} 0 \leq x \leq y \leq 1\\ 0 ,\end{matrix}\right.[/tex]

Therefore

[tex]\int_{0}^{1}\int_{x}^{1}cxy \text{dy dx}=1[/tex]

[tex]c\int_{0}^{1}x\left ( \frac{y^2}{2}\right )_{x}^{1} \text{ dx}=1[/tex]

[tex]\frac{c}{2}\left [ \frac{x^2}{2}-\frac{x^5}{4} \right ]_{0}^1=1[/tex]

[tex]\\\\\frac{c}{2}\left ( \frac{1}{4} \right )=1[/tex]

C=8

Where, the conditional PDF fX|Y(x|0.5)

[tex]f_{x,y}(x|0.5)=\frac{f_{xy}(x,y=0.5)}{f_{y}y}[/tex]

We have

[tex]f_{y}y=8y[\frac{x^2}{2}]_{0}^{y}\[/tex]

[tex]f_{y}y=4y^3[/tex]

[tex]f_{x|y}(x|0.5)=\frac{8x(0.5)}{4(0.5)^3}[/tex]

f_{x|y}(x|0.5)=8x

In conclusion. with the pdf in the form of ax^2

8x comparing ax^b

The value of a = 8 , b = 1 .

Read more about Arithmetic

https://brainly.com/question/22568180


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Let s and t be slack variables.

Initial:

         x           y         s       t       b                Basic

       100        10        1       0     900               s (departing)

       12.5      2.5        0       1      100                t

Z   -2000    -300     0       0       0

   (entering)

Step 1:

         x           y         s           t       b                Basic

         1          1/10     1/100     0      9               x

         0           1        -1/10      1       10              t (departing)

Z        0        -100       20      0       18000

               (entering)

Step 2:

         x       y         s           t         b                Basic

         1       0       1/50     -1/10      8               x

         0      1       -1/10         1        10              y

Z        0      0       10         100     19000

               (entering)

The optimal solution is:

X = 8

Y = 10

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Answers

Answer:

[tex]\frac{dh}{dt} = \frac{216}{1875\pi}[/tex]

Step-by-step explanation:

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[tex]\frac{dh}{dt} = \frac{dv}{dt} * \frac{108}{75\pi h^2}[/tex]

At this point, [tex]h = 10[/tex] and [tex]\frac{dv}{dt} = \frac{8ft^3}{min}[/tex]

So, we have:

[tex]\frac{dh}{dt} = 8 * \frac{108}{75\pi 10^2}[/tex]

[tex]\frac{dh}{dt} = \frac{8 * 108}{75\pi 10^2}[/tex]

[tex]\frac{dh}{dt} = \frac{864}{7500\pi}[/tex]

[tex]\frac{dh}{dt} = \frac{216}{1875\pi}[/tex]

Hence:

The rate is [tex]\frac{216}{1875\pi} ft/min[/tex]

Using implicit differentiation, it is found that the depth of water changes at a rate of -0.0764 feet per minute.

The volume of a cone of radius r and height h is given by:

[tex]V = \frac{\pi r^2h}{3}[/tex]

Applying implicit differentiation, we can find it's rate of change, thus:

[tex]\frac{dV}{dt} = \frac{2\pi rh}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}[/tex]

In this problem:

Height of 12 ft and radius of 10 ft, thus [tex]h = 12, r = 10[/tex].Radius does not change, thus [tex]\frac{dr}{dt} = 0[/tex].Water leaks at a rate of 8 cubic feet per minute, thus [tex]\frac{dV}{dt} = -8[/tex]

Then:

[tex]\frac{dV}{dt} = \frac{2\pi rh}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}[/tex]

[tex]-8 = \frac{\pi (10)^2}{3}\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt} = -\frac{24}{100\pi}[/tex]

[tex]\frac{dh}{dt} = -0.0764[/tex]

The depth of water changes at a rate of -0.0764 feet per minute.

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