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3 years ago

Which transformation represents a reflectiom over the y axis

1 answer:

3 0

A reflection over the Y-axis would be the flipped point or set of points across the y axis while still retaining the same distance from the y-axis.

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Your reading material states that many banks pay interest that is calculated

Step2247 [10]

Carry on... There is not sufficient information to answer

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3 years ago

Hii please help i’ll give brainliest

GarryVolchara [31]

Answer:

-12

Step-by-step explanation:

the opposite of 12 is -12

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3 years ago

you have a bag of 10 marbles. three of them are red ,five are blue, and two are green.what is the probability of not picking a g

Verizon [17]

Answer:

Not probabillity of picking a green marble is 80%

Step-by-step explanation:

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3 years ago

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Joe's bill at grocery store came to $6.08 he paid with a $10 bill and a dime how much change do you get Joe's bill at the grocer

Levart [38]

Answer:

$4.02

Step-by-step explanation:

$10.10 - $6.08 = $4.02

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3 years ago

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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago