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3 years ago

Giving 50 points for both pages + brainlest for first answer

Mathematics

1 answer:

beks73 [17]3 years ago

4 0

Answer:

1) Put a circle on the 1 and make an arrow all the way down to the left of the 1. Then put a circle on the 8 and make an arrow all the way down to the 10.

2) Put a colored in circle on the 6 and make an arrow to right all the way down to the 10. Then put a circle(thats not colored in) on the 4 and make an arrow to the right all the way down to the -10.

3) Put a colored circle on the -6 and 4

Step-by-step explanation:

I hope this helps!!!

Have a wonderful dayyy!!

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Please help asap!!!!

d1i1m1o1n [39]

Answer:

i think you should do multiplication

Step-by-step explanation: I hope that helps.

4 0

2 years ago

Read 2 more answers

What are vertically opposite angles

NeTakaya

When two lines cross, there are four angles at the place where they cross.
Two angles that do NOT share a side are vertical angles. There are two pairs
of vertical angles where the lines cross.

Also, by the way, vertical angles are always equal to each other. It's often
very helpful to know that.

7 0

3 years ago

Read 2 more answers

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 $U\sim N(n,1)$ :