Determine whether the given set S is a

A penny of mass 3.1 g rests on a small 29.1 g block supported by a spinning disk of radius 8.3 cm. The coefficients of friction between block and disk are 0.742 (static) and 0.64 (kinetic) while those for the penny and block are 0.617 (static) and 0.45 (kinetic).

Sketch grapf of f(y)=y for dy\dt = ay+by^2,a>0,b>0,y>0 

Use the method of your choice to factor the trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. 

6y^2 − 23y + 15

Given an sample of data set X which is normal-distributed to $N(\mu =240,\sigma )$, I want to find the ${\displaystyle 95\mathrm{\%}}$ confidence interval of ${\displaystyle max(\mu -200,0)}$. As the ${\displaystyle max(\mu -200,0)}$ cut the normal-distribution curve at the point of 200, the sample of ${\displaystyle max(\mu -200,0)}$ is not more normal distributed. Therefore what is a reasonable ${\displaystyle 95\mathrm{\%}}$ confidence interval?

1. Suppose a fast-food restaurant can expect two customers every 3 minutes, on average. What is the probability that four or fewer customers will enter the restaurant in a 9-minute period?

$y\le 4$

How to solve a cyclic quintic in radicals?
Galois theory tells us that
${\displaystyle \frac{z^{11}-1}{z-1}=z^{10}+z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1}$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ${\displaystyle {\zeta }^1,{\zeta }^2,\dots ,{\zeta }^{10}}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
$\begin{array}{rl}{A}_0& =x_1+x_2+x_3+x_4+x_5\\ A_1& =x_1+\zeta x_2+{\zeta }^2{x}_3+{\zeta }^3{x}_4+{\zeta }^4{x}_5\\ A_2& =x_1+{\zeta }^2{x}_2+{\zeta }^4{x}_3+\zeta x_4+{\zeta }^3{x}_5\\ A_3& =x_1+{\zeta }^3{x}_2+\zeta x_3+{\zeta }^4{x}_4+{\zeta }^2{x}_5\\ A_4& =x_1+{\zeta }^4{x}_2+{\zeta }^3{x}_3+{\zeta }^2{x}_4+\zeta x_5\end{array}$
Once one has ${\displaystyle A_0,\dots ,A_4}$ one easily gets ${\displaystyle x_1,\dots ,x_5}$. It's easy to find ${\displaystyle A_0}$. The point is that ${\displaystyle \tau }$ takes ${\displaystyle A_j}$ to ${\displaystyle {\zeta }^{-j}{A}_j}$ and so takes ${\displaystyle A_j^5}$ to ${\displaystyle A_j^5}$. Thus ${\displaystyle A_j^5}$ can be written down in terms of rationals (if that's your starting field) and powers of ${\displaystyle \zeta }$. Alas, here is where the algebra becomes difficult. The coefficients of powers of ${\displaystyle \zeta }$ in ${\displaystyle A_1^5}$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have ${\displaystyle A_1}$ as a fifth root of a certain explicit complex number. Then one can express the other ${\displaystyle A_j}$ in terms of ${\displaystyle A_1}$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.