Please help me!

Which of the following is not a function B is not right

FinnZ [79.3K]

Answer:

The first one

Step-by-step explanation:

They have 2 outputs for the same input

3 0

3 years ago

Read 2 more answers

Separate the number 41 and the two parts so that the first number is eight more than twice the second number what are the two nu

vesna_86 [32]

The two numbers are 30 and 11

Solution:

Given that we have to separate the number 41 into two parts

Let the second number be "x"

Given that first number is eight more than twice the second number

first number = eight more than twice the second number

first number = 8 + twice the "x"

first number = 8 + 2x

So we can say first number added with second number ends up in 41

first number + second number = 41

8 + 2x + x = 41

8 + 3x = 41

3x = 41 - 8

3x = 33

x = 11

first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30

Thus the two numbers are 30 and 11

7 0

3 years ago

I need help! ill give brainliest!

Lelu [443]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$area = DC \times EG$

$area = 27 \times 16$

$area = 432$

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5 0

3 years ago

Please answer this correctly

Blizzard [7]

Answer:

Commutative Property

Step-by-step explanation:

7 0

3 years ago

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

2 years ago