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3 years ago

12

it takes akios family 2 and 1/2 hours to drive from there home to the beach. it takes his family three times as long to drive to

the mountains as it takes to drive to the beach. How long does it take akios family to drive from there home to the mountain?

1 answer:

dexar [7]3 years ago

6 0

2.5 hours x 3= 7.5 hours

7 and 1/2 hours

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A scale drawing of a playground is shown below.

kolezko [41]

Hm... I think 440 Square feet..?

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4 years ago

If a ants can dig b tunnels in c hours, how many tunnels can b ants dig in a hours?

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9514 1404 393

Answer:

b²/c tunnels

Step-by-step explanation:

The number of tunnels per ant-hour is ...

b/(ac)

Multiplying that by the new number of ant-hours, we have ...

(b/(ac))(ab) = b²/c . . . . tunnels

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3 years ago

A scale drawing of a room is shown below. In the drawing, the room is 73 inches long and

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Step-by-step explanation:

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

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so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago

Point m is the midpoint of kL. M(1,-1) and L(8,-7. What are the coordinates of point k?

Phoenix [80]

Answer: (-6, 5)

<u>Step-by-step explanation:</u>

Use the Midpoint Formula: $\dfrac{(x_k, y_k)+(x_L,y_L)}{2}=(x_m, y_m)$

Separate the x's and y's and solve them individually:

$\dfrac{x_k+x_L}{2}=x_m\qquad \qquad \qquad \qquad\dfrac{y_k+y_L}{2}=y_m\\\\\\\dfrac{x_k+8}{2}=1\qquad \qquad \qquad \qquad \qquad \dfrac{y_k-7}{2}=-1\\\\\\x_k+8=2\qquad \qquad \qquad \qquad \qquad y_k-7=-2\\\\\\x_k\quad =-6 \qquad \qquad \qquad \qquad \qquad y_k\qquad =5$

So, the k-coordinate is (-6, 5)

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3 years ago