Find the exponential model that fits the

Identify each of the following statements as true or false in relation to confidence intervals (CIs). 
Note: 0.5 marks will be taken away for each incorrect answer. The minimum score is 0.

Laplace transform involving step function
$f\left(t\right)=\frac{\sin \left(2t\right)}{e^{2t}}+t·u(t-4)$

The US government is interested in understanding what predicts death rates. They have a set of data that includes the number of deaths in each state, the number of deaths resulting from vehicle accidents (VEHICLE), the number of people dying from diabetes (DIABETES), the number of deaths related to the flu (FLU) and the number of homicide deaths (HOMICIDE).
Your run a regression to predict deaths and get the following output: 
At alpha=0.05, what is indicated by the significance F in this problem? 
A. The regression model does not significantly predict deaths. 
B. The regression model significantly predicts deaths. 
C. Two of the four independent variables significantly predict deaths. 
D. Three of the four independent variables significantly predict deaths. 
$\begin{array}{|cc|}\hline \text{Multiple R}& 0.874642613\\ \text{R Square}& 0.764999351\\ \text{Adjusted R Square}& 0.744564512\\ \text{Standart Error}& 62.97881926\\ \text{Obsevations}& 51\\ \hline\end{array}$

$\text{ANOVA}$

$\begin{array}{|cccccc|}\hline & df& SS& MS& F& SignificanceF\\ \text{Regression}& 4& \mathrm{593934..928}& 148483.732& 37.43603514& 6.36772F-14\\ \text{Residual}& 46& 182451.2571?3966ю331676& & & \\ \text{Total}& 50& 776386.1851& & & \\ \hline\end{array}$

$\begin{array}{|ccccccc|}\hline & Coefficients& StandartError& tStart& P-value& Lower95\mathrm{\%}& Upper95\mathrm{\%}\\ \text{Intecept}& 240.8002285& 51.13392961& 4.709206398& 2.31605E-05& 137.8729666& 343.7274903\\ \text{VEHICLE}& 3.042782981& 1.582326221& 1.922980824& 0.06068593& -0.142274506& 6.227840468\\ \text{DIABETES}& 11.24212265& 1.659066489& 6.776173665& 1.97533E-08& 7.902595017& 14.58165028\\ \text{FLU}& 12.32304584& 2.057012215& 5.990749957& \mathrm{2..9897}E-07& 8.182495007& 16.46359668\\ \text{HOMICIDE}& 1.362128456& 2.113562817& 0.644470297& 0.522471501& -2.892252837& 5.616509749\\ \hline\end{array}$

Use the table for z or t-distribution to determine the value of z or t to construct a confidence interval for a population mean for each of the following combinations of confidence and sample size:

Confidence interval 90% n=37

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.

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.

y^ prime prime -3y^ prime +2y=10 sin x+2 cos 2x . Find a general solution of the following