2 equations, 2 unknowns. Call height H and base B.
From the first sentence:
HB = 216
From the second sentence:
H = ⅔ B
Plug the second into the first and solve for B:
(⅔ B)B = 216
B^2 = 324
B = 18
Plug that back into either of the original 2 equations and solve for H:
H(18) = 216
H = 12
Ok so... W is week. It would be 20+2w. Because we start with the plants hight (20) and them it grows two cm per week, so 2w.
Answer:
The expected cost is 152
Step-by-step explanation:
Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y
if 
and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)
Recall that, by definition
Also, we are given that 
. Recall the following properties of the expected value. If X,Y are random variables, then
Then, using this property we have that ![E[C] = 50+3E[Y]+ 9E[Y^2]](https://tex.z-dn.net/?f=E%5BC%5D%20%3D%2050%2B3E%5BY%5D%2B%209E%5BY%5E2%5D)
.
Thus, we must calculate E[Y] and E[Y^2].
Using the definition, we get that
![E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%20%5Cint_%7B1%7D%5E%7B5%7D%5Cfrac%7By%7D%7B4%7D%20dy%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft%5Cfrac%7By%5E2%7D%7B2%7D%5Cright%7C_%7B1%7D%5E%7B5%7D%20%3D%20%5Cfrac%7B25%7D%7B8%7D-%5Cfrac%7B1%7D%7B8%7D%20%3D%203)
![E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}](https://tex.z-dn.net/?f=E%5BY%5E2%5D%20%3D%20%5Cint_%7B1%7D%5E%7B5%7D%5Cfrac%7By%5E2%7D%7B4%7D%20dy%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft%5Cfrac%7By%5E3%7D%7B3%7D%5Cright%7C_%7B1%7D%5E%7B5%7D%20%3D%20%5Cfrac%7B125%7D%7B12%7D-%5Cfrac%7B1%7D%7B12%7D%20%3D%20%5Cfrac%7B31%7D%7B3%7D)
Then
30 × 30 + 50 × 50 = 3400
square root of 3400 is 58.31 feet.
58.31 feet is the angle of elevation of the kite.
