What is 1,0000 +1,00000 plz help

According to the U.S. Census Bureau (2016), a substantial pay gap still exists. The average American man, with an advanced colle

Leokris [45]

Answer:

The difference in earnings, over a 30-year career, for men vs women, is $1,200,150

Step-by-step explanation:

Per year.

The average man earns $90,761.

The average woman earns $50,756

So, per year, the difference is:

90,761 - 50,756 = 40,005

Over 30 years:

30*40,005 = 1,200,150

The difference in earnings, over a 30-year career, for men vs women, is $1,200,150

0 0

3 years ago

What step do u preform first when evaluating the expression 3 + (18 - 6) + 20 divided by 4

Elena L [17]

Answer:

Step-by-step explanation:

=3+(18-6)+20÷4

=3+12+20÷4

=3+12+5

=20

use BODMAS rule (Bracket of Division, Multiplication, Addition, and Subtraction)

5 0

2 years ago

Read 2 more answers

Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.

Elden [556K]

Answer:

( 6(x)² + 12(x) - 90 ) ft³

Explanation:

deep end water: 2(x)² + 12(x) + 10

shallow end water: 4(x)² - 100

addition problem: 2(x)² + 12(x) + 10 + 4(x)² - 100

total volume of water:

2(x)² + 12(x) + 10 + 4(x)² - 100

2(x)² + 4(x)² + 12(x) + 10 - 100

6(x)² + 12(x) - 90

7 0

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