Answer:
-5 or 5 depending on how you write the simplified form
Step-by-step explanation:

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The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
1/a^b = a^-b
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Some folks like their simplified form to contain only positive exponents. Other folks avoid unnecessary fractions. Choose the form that satisfies your grader.
6 minutes/60minutes = 0.1 hrs
45miles/0.1hour = 450 miles per hour or mph
If
is the cumulative distribution function for
, then

Then the probability density function for
is
:

The
th moment of
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](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5Enf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bn-1%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Let
, so that
and
:
![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](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu%7De%5E%7B-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du)
Complete the square in the exponent:

![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](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B%5Cfrac12%28n%5E2-%28u-n%29%5E2%29%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac%7Be%5E%7B%5Cfrac12n%5E2%7D%7D%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du)
But
is exactly the PDF of a normal distribution with mean
and variance 1; in other words, the 0th moment of a random variable
:
![E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1](https://tex.z-dn.net/?f=E%5BU%5E0%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du%3D1)
so we end up with
![E[Y^n]=e^{\frac12n^2}](https://tex.z-dn.net/?f=E%5BY%5En%5D%3De%5E%7B%5Cfrac12n%5E2%7D)
Hope this helps but I think it is A.)addition property of equality.
Answer:

Step-by-step explanation:
The inverse of a slope is the exact opposite slope of the original. I'll show a function below where you can easily solve future problems if you need to.

m is the slope, so for the equation above, you simply insert the original slope, being 1, in the equation.

So the inverse slope would be -1.