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3 years ago

A rocket is launched from atop a 56-foot cliff with an initial velocity of 135ft/s.

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

4 0

Answer:

16.56.135

Step-by-step explanation:

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Geometry math question no Guessing and Please show work

Vera_Pavlovna [14]

IN ORDER TO FIND THE ANSWER:

Add up two of the values. If they are greater than the third, the lengths can make up a triangle.

3 + 9 = 12
12 > 14? No

3 + 5 = 8
8 > 7? Yes

1 + 2 = 3
3 > 3? No

4 + 4 = 8
8 > 8? No

The answer would be B.

5 0

3 years ago

Each year, roughly 10^6 computer programmers each make about $ 10^5.

Xelga [282]

Answer:

100,000,000,000

Step-by-step explanation:

1 000 000 times 100 000 = 100,000,000,000

5 0

3 years ago

The product of a constant factor 15 and a 2-term factor x+4

dezoksy [38]

The product of 15 and $x+4$ is $15(x+4) = 15x+60$ . This is a linear binomial.

$4x^3-2x-9-(2x^3)$ is a cubic trinomial. In your original question, I'm not sure what the $5x+3$ means.

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