Answer:
i think bca would be the same
Step-by-step explanation:
Answer:
11
Step-by-step explanation:
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Please ignore the random characters above
Glad to help you =)
Answer:

Step-by-step explanation:
Given
--- interval
Required
The probability density of the volume of the cube
The volume of a cube is:

For a uniform distribution, we have:

and

implies that:

So, we have:

Solve


Recall that:

Make x the subject

So, the cumulative density is:

becomes

The CDF is:

Integrate
![F(x) = [v]\limits^{v^\frac{1}{3}}_9](https://tex.z-dn.net/?f=F%28x%29%20%3D%20%5Bv%5D%5Climits%5E%7Bv%5E%5Cfrac%7B1%7D%7B3%7D%7D_9)
Expand

The density function of the volume F(v) is:

Differentiate F(x) to give:




So:

9514 1404 393
Answer:
- central: XWR, VWU
- inscribed: VST
Step-by-step explanation:
Point W is the center of the circle. Any angle with W as its vertex is a central angle:
angles XWR and VWU are central angles
Any angle with its vertex on the circle and rays that intersect the circle is an inscribed angle.
angle VST is an inscribed angle
Answer:
12
Step-by-step explanation:
Use the pythagorean theorem (
). A and b are the two legs of the triangle, and c is ALWAYS the hypotenuse. Plug in the values for a and c




