Rom

\(|z^4| = |z|^4\\ |z^4|^2 = \dfrac{64^2}{25} + \dfrac{48^2}{25} = 256\\ |z^4|=16\\ |z| = \sqrt[4]{16}=2\)

I checked it over first 10 million integers.

It appears to transition from 15048 at 19998 to 15049 at 19999

and remain there.

\(u^3 -3uv^2 + 3u^2 v - v^3 = (u-v)(u^2 + 4uv+v^2) = 259-286= -27\)

\(\text{This has to be factored as }\\ (u-v)=9\\ (u^2+4uv+v^2)=-3\\ \text{in order to match the equation, and further must be that }\\ u=7, v=-2\\ u^2+v^2 = 49+4 = 53\)

appears to be 15049

\(z+\dfrac 1 z = z^2 + \dfrac{1}{z^2}\\ z^4-z^3-z+1=0\)

\(\text{A quick eyeball solution is }z=1\\ \text{So let's divide through by }(z-1)\\ z^3-1 = 0\\ \text{This also clearly has a solution at }z=1 \text{ so divide again}\\ z^2+z+1 = 0\\ \text{This is just quadratic so we can apply the quadratic formula}\\ z = \dfrac{-1\pm \sqrt{1-4}}{2} = -\dfrac 1 2 \pm i\sqrt{3}\\ \text{So in total we have}\\ z = 1, ~z= -\dfrac 1 2 + \dfrac{i\sqrt{3}}{2}, ~z = -\dfrac 1 2 - \dfrac{i\sqrt{3}}{2}\)

\(w = 1 + \dfrac 1 1 = 2\\ -\dfrac 1 2 \pm i \sqrt{3} = e^{\pm i\frac{2\pi}{3}}\\ e^{\pm i\frac{2\pi}{3}}+e^{\mp i\frac{2\pi}{3}} = 2\cos\left(\dfrac{2\pi}{3}\right) = -1 \)

\(p=\Phi\left(\dfrac{276-300}{53}\right) =\\ \Phi(-0.45283)= \text{ (using software or lookup table)}\\ 0.325 =32.5\% \)

\(N+1x = 97\\ N+5x = 265\)

\(\text{Subtract eq 1 from eq 2}\\ 4x = 168\\ x = \$42\\ N = 97-x = 97-42 = \$55\)

just plug it in?

\(f(-6) = \dfrac{100}{-10+e^{-0.1 \cdot (-6)}} = \dfrac{100}{-10 +e^{0.6}}\\ \text{dump that into a calculator and round the answer to the nearest tenth}\)

I'm just curious. 

This is about the most basic thing you can do when learning statistics.  Use the z-score table to 

look up the associated probability. 

Did no one show you how to do this?  Do you not have a table available? (hint they are all over the web)

Do you not care at all that you don't learn how to do this?  How's that going to work on a test?