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:
Test the choices !
Pick an even number, and see what each choice does to it.
Let's start with, say, 6 .
We'll try each choice, and see which one produces an odd number:
a). 6 to the 2nd power. . . . . . 6 x 6 = 36. That's not an odd number.
b). 6 + 3 = 9 This could be it. 9 is odd. We'll save this one.
c). 3·6 = 18. That's not an odd number.
d). 6/3 = 2. That's not an odd number.
The only one that gave us an odd number is (b).
Answer:
Step-by-step explanation:
8 in × (1 ft)/(12 in) = ⅔ ft
36 ft × 58 ft × ⅔ ft = 1392 ft³
1392 ft³ × (1 yd³)/(27 ft³) ≅ 52 yd³
52 yd³ × $123/yd³ = $6396
Answer:
2311
Step-by-step explanation:
311,225,823,387,830,850,069
Oof
