Answer:
I think the author is meaning to tell you to use your imagination to think about what will happen when you enter. You don’t know until you through it.
Step-by-step explanation:
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:

Answer:
Area covered by the fences will be 16.1 unit²
Step-by-step explanation:
Let the first parabola is represented by the function f(x) = 6x²
and second parabola by g(x) = x² + 9
point of intersection of the graphs will be determined when f(x) = g(x)
6x² = x² + 9
5x² = 9
x² = 1.8
x = ± 1.34
Now we will find the area between these curves drawn on the graph.
Area = ![\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx](https://tex.z-dn.net/?f=%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5Bf%28x%29-g%28x%29%5Ddx%3D%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5B6x%5E%7B2%7D-%28x%5E%7B2%7D%2B9%29%5Ddx)
= 
= ![[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7Dx%5E%7B3%7D-9x%5D_%7B-1.34%7D%5E%7B1.34%7D)
= ![[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7D%28-1.34%29%5E%7B3%7D-9%28-1.34%29-%5Cfrac%7B5%7D%7B3%7D%281.34%29%5E%7B3%7D%2B9%281.34%29%5D)
= ![[-4.01+12.06-4.01+12.06]](https://tex.z-dn.net/?f=%5B-4.01%2B12.06-4.01%2B12.06%5D)
= 16.1 unit²
Answer:
-4.5
Step-by-step explanation:





Answer: no tommyinnit is swag let him rock dude
Step-by-step explanation: cause i said so