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3 years ago

Solve the following equation: 7/8 = -7

Mathematics

1 answer:

guajiro [1.7K]3 years ago

6 0

Answer:

lets just say -8

Step-by-step explanation:

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Use the elimination method to solve the system of equations. Choose the correct ordered pair.

Mandarinka [93]

Answer:

based on yahoo, cause my iq is like 10...the answer is B

Step-by-step explanation:

6 0

3 years ago

1. Kyle and John are twins who decided to plan an imaginary trip that started with cliff jumping. There were two different spots

Vikki [24]

Answer:

a) Δx = -35ft

b) Δx = -25ft

c) Kyle traveled 10ft more than John

Step-by-step explanation:

We define a coordinate reference system() in which y = 0 corresponds to the water surface.

a) The initial position of Kyle in our coordinate reference system is:

$x_{o} = 20 ft$

and his final position is:

$x_{f} = -15 ft$

Therefore, he traveled

Δx = $x_{f} - x_{o} = -15 - 20 = -35 ft$

b) The initial position of John in our coordinate reference system is:

$x_{o} = 15 ft$

and his final position is:

$x_{f} = -10 ft$

Therefore, he traveled

Δx = $x_{f} - x_{o} = -10 - 15 = -25 ft$

c) Then, Kyle traveled 10ft more than John

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