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tatiyna
3 years ago
11

The blue dot is at what value on the number line?

Mathematics
1 answer:
iren2701 [21]3 years ago
6 0

Answer:

the blue dot is at -19 on the number line

Step-by-step explanation:

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Figure ABCD has verticies A(-4, 1) B(2, 1) C(2, -5) D(-4-3). What was the area of Figure ABCD.
JulijaS [17]

area: 18 units^2

Step-by-step explanation:

the shape is a quadrilateral, with one slanted side, so I separated the shape into a rectangle and triangle, and and calculated their area respectively, then added the products up. please correct me if I'm wrong. hope this helped. :)

8 0
3 years ago
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HELP - Wesley subscribes to a telecommunication company's move package. The plan includes a basic monthly charge of $40 and an a
boyakko [2]

Answer:

(month x 40) + (movies x C) or 40 + (M x C)     :/

Step-by-step explanation:

8 0
3 years ago
5. (Total 16 points)
Mila [183]
Its nuts thats what it is 

4 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 nth 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
3 years ago
The surface area of 3ft, 7ft, 2ft
Xelga [282]

Answer: 82 ft squared

Step-by-step explanation: The surface area is 2( wl + hl + hw)

so the width is 2 the length is 7 and the height is 3

so it's 2( 14 + 21 + 6) = 2(41) = 82

6 0
3 years ago
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