Find the area of this figure.

Mathematics

1 answer:

jeyben [28]2 years ago

3 0

180.48 ft^2

Hope it’s right
Best luck with your studying

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How do you solve this

brilliants [131]

First, ask somebody else like your teacher, also use your brain

5 0

2 years ago

Find the volume of the composite figure. please!

Irina-Kira [14]

Volume= Length * width ( aka Base ) * height

Break up the figure into easier figures, a small square and a bigger square.

Small square- 3*2*1 = 6 cm

Large Square- 7*6*1 = 42 cm

42 + 6 = 48 cm. But wait! You have to take one more step, which is minus-ing 6 from 48. Why? Notice that there is a little area where a side of the small square meets the bigger square. That little area is worth 3 cm ( length is 3, height is 1 cm ) times 2 ( 3 cm is one side, another 3 cm is the other side from the other square ) = 6 cm.

Your total answer should be 42. ( or 48, if your teacher doesn't count the area where the squares meet/join together ).

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