Answer:
![P(Cube \ge 5\ and\ Spin[A]) = \frac{1}{24}](https://tex.z-dn.net/?f=P%28Cube%20%5Cge%205%5C%20and%5C%20Spin%5BA%5D%29%20%3D%20%5Cfrac%7B1%7D%7B24%7D)
Step-by-step explanation:
Given
See attachment for spinner
Required
![P(Cube \ge 5\ and\ Spin[A])](https://tex.z-dn.net/?f=P%28Cube%20%5Cge%205%5C%20and%5C%20Spin%5BA%5D%29)
On a number cube, we have:
---- i.e. 2 outcomes
So, the probability is:
On the spinner, we have:
![Spin [A] = \{A\}](https://tex.z-dn.net/?f=Spin%20%5BA%5D%20%3D%20%5C%7BA%5C%7D)
---- i.e. 1 outcomes
So, the probability is:
![P(Spin[A]) = \frac{n(Spin[A])}{n(Spinner)}](https://tex.z-dn.net/?f=P%28Spin%5BA%5D%29%20%3D%20%5Cfrac%7Bn%28Spin%5BA%5D%29%7D%7Bn%28Spinner%29%7D)
![P(Spin[A]) = \frac{1}{8}](https://tex.z-dn.net/?f=P%28Spin%5BA%5D%29%20%3D%20%5Cfrac%7B1%7D%7B8%7D)
is calculated as thus:
![P(Cube \ge 5\ and\ Spin[A]) = P(Cube \ge 5) * P(Spin[A])](https://tex.z-dn.net/?f=P%28Cube%20%5Cge%205%5C%20and%5C%20Spin%5BA%5D%29%20%3D%20P%28Cube%20%5Cge%205%29%20%2A%20P%28Spin%5BA%5D%29)
![P(Cube \ge 5\ and\ Spin[A]) = \frac{1}{3} * \frac{1}{8}](https://tex.z-dn.net/?f=P%28Cube%20%5Cge%205%5C%20and%5C%20Spin%5BA%5D%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%2A%20%5Cfrac%7B1%7D%7B8%7D)
Please be clear with the expressions.
Here we assume
f(x)=(1/3) sqrt(x)
A scale factor is a multiplier applied to a parent function.
If the parent function is f(x), and
g(x)=a*f(x), then a is a scale factor.
Assuming the parent function is sqrt(x), and f(x)=(1/3) sqrt(x)
then the scale factor is (1/3).
[note]
if f(x) is a cube root, then write
f(x)=cube root(x), or f(x)=x^(1/3).
the exponentiation sign ^ is very useful.
I’m pretty sure you can first do 10% at a time which you can round to 13 which is 3 so 3 plus 3 then add tax
Answer:
11.547 ft wide by 5.774 ft high
769.800 ft³ capacity
Step-by-step explanation:
Volume is maximized for a given area by having the area of a pair of opposite sides equal the area of the bottom. That means the overall area of the container is 3 times the area of the bottom. Then the square bottom will have a width of ...
w = √(400/3) ≈ 11.547 . . . feet
The height is half that, so is ...
h = w/2 = 11.547/2 ≈ 5.774 . . . feet
The capacity is then ...
w²h = (11.547 ft)²(5.774 ft) = 769.800 ft³
The container is 11.547 ft wide by 5.774 ft high. It has a capacity of 769.800 cubic feet.
_____
You want to maximize w^2h subject to w^2 + 4wh = 400. Solving the constraint equation for h, we get h = (400 -w^2)/(4w) and the volume we want to maximize can be written as ...
V = w(400-w^2)/4
This will be an extreme when dV/dw = 0, so we want to solve ...
dV/dw = 0 = 100 -(3/4)w^2
w^2 = 400/3
w = √(400/3) . . . . . as above
The answer will be 1.09. Hope this will help you!
