How many 1/4 cups could you measure out of 2 and 2/4 cups

Mathematics

1 answer:

maksim [4K]3 years ago

5 0

2 and 2/4 = 10/4

10/4 ÷ 1/4 = 10/4 × 4/1 = 10 cups

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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

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