According to one survey in India, 75% of

Finding the extrema of a functional (calculus of variations)
${\displaystyle I\left(y\right)={\int }_0^1{y}^{\prime 2}-y^2+2xy dx }$
Explanation:
$\begin{array}{rl}\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}& =2y^{\text{'}}⟹\frac{d}{dx}\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{y}^{\text{'}}}\right)=2y^{\text{'}\text{'}}\\ \frac{\mathrm{\partial }F}{\mathrm{\partial }y}& =-2y+2x\end{array}$
Which gives the Euler equation ${\displaystyle 2y{ }^{″}+2y-2x=0}$ (I think, unless I messed up my math).
Ok now I need to try to find solutions to this differential equation, however, I only know how to solve linear second order DE's. The x term is throwing me off and I am not sure how to solve this.
I know ${\displaystyle y\left(x\right)=x}$ is a solution by inspection, but inspection is a poor man's approach to solving DE's.

Simplify (condense) the following expression.

$7{\log }_{5}p-{\log }_{5}q$

you throw a glob of putty straight up toward the ceiling which is 3.60m above the point where the putty leave you hand the initial speed of the putty as it leaves your hand is 9.50m/s.

I calculated ${\displaystyle {\sin 75}^{\circ }}$ as ${\displaystyle \frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}}$, but the answer is ${\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}}$. What went wrong?
I calculated the exact value of ${\displaystyle {\sin 75}^{\circ }}$ as follows:
${\displaystyle {\sin 75}^{\circ }=\sin ({30}^{\circ }+{45}^{\circ })}$
$=\sin 30°\cos 45°+\cos 30°\sin 45°$
$=\frac{1}{2}·\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}$
$=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$
The actual answer is
${\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}}$

The data given to the right includes data from

46

​candies, and

8

of them are red. The company that makes the candy claims that

31​%

of its candies are red. Use the sample data to construct a

90​%

confidence interval estimate of the percentage of red candies. What do you conclude about the claim of

31​%?

Parameters for which a quadratic matrix equation has a unique solution
Let function $f:{\mathbf{R}}^{n\times n}\to \mathbf{R}$ be defined by
${\displaystyle f\left(X\right)=a^{T}Xb+c^{T}XEXd}$
where a, b, c, ${\displaystyle d\in {\mathbb{R}}^n}$ and ${\displaystyle n\times n}$ matrix E is symmetric and positive semidefinite.
Under what conditions on a, b, c, d and E is there a unique matrix X such that ${\displaystyle f\left(X\right)=0}$?
This cannot always be true, of course. For instance, if ${\displaystyle a=b=c=d=0}$ then any matrix X will satisfy ${\displaystyle f\left(X\right)=0}$.

You measure weekly sales. You can assume that weekly sales are normally distributed with mean 60 and standard deviation 18, and that weekly sales are independent across weeks.

What is the probability that AVERAGE WEEKLY sales over the next 4 weeks are below 66 in 4 decimal positions?