PLEASE HELP ASAP ! Pictures are linked below thank you

Mathematics

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

see below

Step-by-step explanation:

Step A is wrong because when taking terms (x - 6) to the left side these signs should change but they did not change in 1st step

John set the equation equal to zero incorrectly

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A system has one each of two different types of components in joint operation. Let X and Y denote the random lengths of the live

RSB [31]

Answer:

The value of E(Y/X) is 2.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

$f(x,y)=\left \{ {{\dfrac{x}{8}e^{-\dfrac{x+y}{2}} \,\,\,\,0 \leq x,0\leq y \atop {0}} \right.$

So the marginal function for X is given as

$f_x=\int\limits^{\infty}_0 {f(x,y)} \, dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x+y}{2} }dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x}{2}}e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{8}e^{-\frac{x}{2}}\int\limits^{\infty}_0e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{4}e^{-\frac{x}{2}}$

Now

$f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }$

Now the value of E(Y/X) is given as

$E(Y/X)=\int\limits^{\infty}_0 {yf_{Y/X}} \, dy \\E(Y/X)=\int\limits^{\infty}_0 {y\dfrac{1}{2}e^{-\frac{y}{2} } }\, dy\\E(Y/X)=\dfrac{1}{2}\dfrac{\sqrt{2}}{(\dfrac{1}{2})^2}=2$

So the value of E(Y/X) is 2.

3 0

3 years ago

How do I solve this I don't understand it

Free_Kalibri [48]

Here’s how u do it ^

7 0

3 years ago

Find the area round to two decimal places as needed

Mademuasel [1]

To find the area of an obtuse triangle you have to multiply the base of the triangle by the vertical height and divide the result by 2 following the formula: