Expert Help with High School Geometry

Volume of a Truncated Right Prism with generic base convex polygon
I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}(h_1+h_2+h_3)$, where A is the area of the base and $h_i$ is the height of vertex i of the top face. Or, in other words, to find the volume you "flatten" the top face by finding the average height of its 3 vertices, and then the volume is just that of a regular prism.
My question is: for a generic base convex polygon, does the same argument apply? Is the volume equal to $\frac{A}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{h}_i$, where n is the number of vertices of the base/top? Is there a proof for this anywhere?
My intuition says it's true, but what if the base shape was like a really long kite; wouldn't the height of the vertex at the bottom tip of the kite skew the average height of the top face?

If I have a circle that has ten dots on it and I have to make as many polygons using those dots, the answer should be 10. Connecting each dot starting with triangle, to square, all the way to the max of a 10-sided polygon should give me only 10 polygons that I can make. Am I missing something here?

Geometric distribution and Conditional probability problem
I got this problem in my book.. I will directly go the last part which I could not solve.
$P(goodcondition)=0.91854$
$P(badcondition)=0.08146$
a random package check is checked by a company till there is a package in a bad condition .
1. What is the probability that the company will check exactly 4 packages?
2. It is known that the first 3 packages that were checked are in good condition , what is the probability that more than 8 packages will be checked(edit: the answer should be 0.654).

Indirect proof $gcd(9k+4,2k+1)=1$ $\forall k\in Z$

Finding volumes - when to use double integrals and triple integrals?
This is not a technical question at all, but I'm quite confused about what should I use to compute volumes in ${\mathbb{R}}^3$ with integration.
I've read somewhere that a double integral gets the volume swapping across the x and y axis while a triple integral just integrate the whole thing at once, how accurate is this?. Can a volume expressed by a double integral be expressed by a triple integral?, and a triple can always be expressed by a double?

Distance between (-6,3,4) and (-1,1,3)?

Geometric Probability. Sphere
Suppose a sphere with radius R. Find the probability of the event such that n selected points of the sphere are within the distance of $r=\frac{R}{2}$ from the center of the sphere. The points are uniformly distributed and selected independently.

Distance between (-7,8) and (3,5)?

Volume of a parallelepiped, given 8 vertices
Given the eight vertices (0,0,0), (3,0,0), (0,5,1), (3,5,1), (2,0,5), (5,0,5), (2,5,6), and (5,5,6), find the volume of the parallelepiped.
I'm having trouble finding the 1 vertex and 3 vectors needed to find the volume. The closest four vertexes I found so far are (0,0,0),(3,0,0),(0,5,1),(3,5,1)...is using those four vertexes correct? Any starting hints to point me in the right direction?

What is the distance between (18,5) and (7,4)?

Finding volume of solid
Suppose that a solid is formed in such a way that each cross section perpendicular to the x-axis, for $0\le <!--\; \le \; -->x\le <!--\; \le \; -->1$, is a disk, a diameter of which goes from the x-axis out to the curve $y=\sqrt{x}$.
Find the volume of the solid.
For this I use the disk formula. So $\mathrm{\pi <!--\; \pi \; -->}{\displaystyle {\int <!--\; \int \; -->}_0^1(\sqrt{x}{)}^{2}dx.}$.
When I do this, I get ${\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{5}}$. The answer is ${\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{8}}$. What am I doing wrong?

Finding the probability distribution of the sum of geometric distributions?
X and Y are both geometric distributions with success p. What it the $Pr(X+Y=n)$
Would I use a convolution with a sum for this? Do I need to define a third random variable?

Distance between (4,3,1) and (3,6,2)?

$\mathrm{△<!--\; △\; -->}ABC$ has D a point on side BC. Points E and F are on sides AB and AC such that DE and DF are the angle bisectors of $\mathrm{\angle <!--\; \angle \; -->}ADB$ and $\mathrm{\angle <!--\; \angle \; -->}ADC$ respectively. Find the value of $\frac{AE}{EB}\ast <!--\; \ast \; -->\frac{BD}{DC}\ast <!--\; \ast \; -->\frac{CF}{FA}$

I am trying to prove the inscribed angle theorem using vectors. I fixed the dots $A=(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$, and $B=(\cos <!--\; \; -->\mathrm{\phi <!--\; \phi \; -->},\sin <!--\; \; -->\mathrm{\phi <!--\; \phi \; -->})$, and I took another point $C=(\cos <!--\; \; -->\mathrm{\psi <!--\; \psi \; -->},\sin <!--\; \; -->\mathrm{\psi <!--\; \psi \; -->})$ in the biggest arc AB.
My idea was to calculate ${\displaystyle \frac{⟨<!--\; ⟨\; -->A-<!--\; -\; -->C,B-<!--\; -\; -->C⟩<!--\; ⟩\; -->}{|<!--\; |\; -->A-<!--\; -\; -->B|<!--\; |\; -->|<!--\; |\; -->B-<!--\; -\; -->C|<!--\; |\; -->}}$, what according to my calculations is ${\displaystyle \frac{1+\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\phi <!--\; \phi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\psi <!--\; \psi \; -->}-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})}{2\sqrt{1+\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\phi <!--\; \phi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->})}}}.$.
My main difficulty here is the square root, which I can't get rid of. Does someone know how to proceed from here?Or maybe to solve the problem with vectors with a different approach?

Geometric Interpretation of Independence of Random Variables on the Standard Probability Space
Let $([0,1],\mathcal{R}\cap <!--\; \cap \; -->[0,1],\mathrm{\lambda <!--\; \lambda \; -->})$ be the standard probability space on [0,1] with $\mathrm{\lambda <!--\; \lambda \; -->}$, the Lebesgue measure. What do independent random variables "look like" in this space? Is there an easy geometric approach to creating independent functions on this space?

Continuous/probability version of arithmetic/geometric mean inequality
If one has a random variable X, described by a finite probability distribution with equally likely possible values $x_1,\dots <!--\; \dots \; -->,x_n$, then the inequality: $E[\log <!--\; \; -->X]\le <!--\; \le \; -->\log <!--\; \; -->(E[X])$ is a reformulation of the arithmetic/geometric mean inequality.
And a necessary and sufficient condition for equality (maybe only if the distribution is concentrated at one point, for example)?

O is intersection of diagonals of the square ABCD. If M and N are midpoints of OB and CD respectively ,then $\mathrm{\angle <!--\; \angle \; -->}ANM=?$

Finding the volume of a solid region
I'm trying to find the volume of the solid region inside the sphere $x^2+y^2+z^2=4$, and the upper nappe of the cone $z^2=3x^2+3y^2$ (I only have to set up the triple integral itself, not evaluate it).
So, I need to do this in cylindrical coordinates. Thus, for the r value, since $z^2+y^2=4$, $r=2$. Thus, $0\le <!--\; \le \; -->r\le <!--\; \le \; -->2$.
Next, I need to determine the value of z. I've considered the fact that $y=\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$, and $x=\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$. Thus, $z^2=3({\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))+3({\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))$. This is equivalent to $z^2=3$. Thus, $z=\sqrt{3}$. This would be the lower limit, because it came from the equation of the cone. For the upper boundary of z, I'm not completely sure. Any help with this would be greatly appreciated.
Finally, for $\mathrm{\theta <!--\; \theta \; -->}$, I think it would be 0 to $2\mathrm{\pi <!--\; \pi \; -->}$, since it is the solid region inside a sphere.
I'm not extremely confident in my abilities yet, and I don't want to lead myself astray. Any help/corrections would be greatly appreciated!

Distance between (-6,-6) and (3,5)?