PLEASE HELP ME ANSWER THIS BECAUSE IM NOT SMART

What is the x intercept for g(x) = 3|x - 2| - 2.

kogti [31]

Answer: g(x)= 31x-21-2

Step-by-step explanation:

21-2=19

31x=19

divide that by 31

the answer would be 1.63

3 0

2 years ago

Please hurry I need to pass this class

Dvinal [7]

Answer:

$\displaystyle m=\frac{-3}{4}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Reading a Cartesian plane
Coordinates (x, y)
Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point A(0, 8)

Point B(4, 5)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [Slope Formula]: $\displaystyle m=\frac{5-8}{4-0}$
[Fraction] Subtract: $\displaystyle m=\frac{-3}{4}$

4 0

2 years ago

Which relation is a function?

yKpoI14uk [10]

Answer:

C

Step-by-step explanation:

if you don't have from same x different outcomes is a function

think that x is a day you count votes, you can not get for the same day different y number of votes ( something is fishy then)

A. not a function because (6, -3) and (6, 3)

B. not a function because (4,7) and (4,2)

C. is a function

D. not a function because (2, -1) (2,1) (2,3)

4 0

2 years ago

I neee #14 PLEASE! Thanks! <3

Ksenya-84 [330]

I hope this helps you

5 0

2 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