The function h is defined by h(x)=4+3x over 5+x Find h(4z)

Mathematics

1 answer:

Jobisdone [24]3 years ago

5 0

Answer:

h(4z)=4+12z / 5+4z

Step-by-step explanation:

The function h is defined by h(x)=4+3x over 5+x

This can be written as

h(x)=4+3x / 5+x

The value of the function at any time depends on the value of x. x is the independent variable. The variable,x can take any value at any time.

To find h(4z), it means that x is taking the value of 4z.

We will substitute 4z in the place of x in the function. This becomes

h(4z)=4+3×4z / 5+4z

h(4z)=4+12z / 5+4z

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Step-by-step explanation:

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

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

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

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