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3 years ago

14

The crafting club sold art calendars for a recent fundraiser. The table below lists the number of calendars sold by each member

the first two weeks of the sale.

1 answer:

solniwko [45]3 years ago

3 0

Answer:

there is no table below, you have to add a picture

Step-by-step explanation:

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At the craft store, Tara bought 12 yards of fabric to make tote bags. Each bag requires 1.25 yards of fabric. If Tara makes 5 to

matrenka [14]

Answer:

5.75 yds left

Step-by-step explanation:

1.25 yds per bag*5 bags =6.25 yds used

She had 12 yds

12-6.25 =5.75

She has 5.75 yds left

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2 years ago

Please help me out! And show your work :)) <3

8_murik_8 [283]

Answer:

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Step-by-step explanation:

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3 years ago

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Evaluate. Show all work<br> 51-(21-5)x2

masya89 [10]

Answer:

19

Step-by-step explanation:

51-(16)x2

16x2=32

51-32=19

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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

Sloving two-step equation

Sedaia [141]

Answer:

what

Step-by-step explanation:

can i please have brainlyest have a good day

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3 years ago