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astra-53 [7]
1 year ago
5

Represent the following expressions as a power of the number a where a ‡0.

Mathematics
1 answer:
Yuri [45]1 year ago
3 0

Properties we are going to use :
1. (a^m)^n = a^mn
2. a^m : a^n = a^(m-n)

((a^-2)^-1)^-1 = (a^(-2 x -1))^-1 = (a^2)^-1 = a^-2

(a : a^-1)^2 = (a^1 - (-1))^2 = (a^2)^2 = a^4

Therefore, the expression is simplified down to a^-2 : a^4

a^-2 : a^4 = a^(-2 - 4) = a^-6.

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If y=2 when x=3. What is the value of y when x=9?
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Because x was *3 you had to y*3 which is 
y=6

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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
Can anyone help me out on this not too sure on this
Paraphin [41]

Answer:

y = 0.1x - 4

Step-by-step explanation:

Given the following data;

Points on the x-axis (x1, x2) = (-3, 7)

Points on the y-axis (y1, y2) = (-7, -6)

First of all, we would find the slope;

Mathematically, slope is given by the formula;

Slope = \frac{Change \; in \; y \; axis}{Change \; in \; x \; axis}

Slope, m = \frac {y_{2} - y_{1}}{x_{2} - x_{1}}

Substituting into the equation, we have;

Slope = (-6 - (-7))/(7 - (-3))

Slope = (-6 + 7)/(7 + 3)

Slope = 1/10

Slope = 0.1

Next, we would write the equation using the formula;

y - y1 = m(x - x1)

y - (-7) = 0.1(x - (-3))

y + 7 = 0.1(x + 3)

y + 7 = 0.1x + 3

y = 0.1x + 3 - 7

y = 0.1x - 4

5 0
3 years ago
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