We are given the functions:
<span>S (p) = 40 + 0.008 p^3 --->
1</span>
<span>D (p) = 200 – 0.16 p^2 --->
2</span>
T o find for the price in which the price of supply equals
demand, all we have to do is to equate the two equations, equation 1 and 2, and
calculate for the value of p, therefore:
S (p) = D (p)
40 + 0.008 p^3 = 200 – 0.16 p^2
0.008 p^3 + 0.16 p^2 = 160
p^3 + 20 p^2 = 20,000
p^3 + 20 p^2 – 20,000 = 0
Calculating for the roots using the calculator gives us:
p = 21.86, -20.93±21.84i
Since price cannot be imaginary therefore:
p = 21.86
Answer:
-8
Step-by-step explanation:
f(x) = 3x + 1
f(-3) = 3(-3) + 1
F(-3)=-9+1
f(-3)= -8
Answer: 13.5 Okay! Here's the method count the legs of the right triangle
The formula we'll use will be
A^2 + B^2 = C^2
In this case we're counting by twos
The base is 11 so we times it by itself =110
The leg is 8.5 so we going to times itself to make 72.25 add those together so 110+ 72.25 = 182.25 then we \|-----
182.25
Then you have got ur answer of 13.5
Step-by-step explanation:
Answer:
I think it's 300
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given
Solving (a):
Find k
To solve for k, we use the definition of joint probability function:
Where
Substitute values for the interval of x and y respectively
So, we have:
Isolate k
Integrate y, leave x:
![k \int\limits^2_{0} y {dx} \, [0,x/2]= 1](https://tex.z-dn.net/?f=k%20%5Cint%5Climits%5E2_%7B0%7D%20y%20%7Bdx%7D%20%5C%2C%20%5B0%2Cx%2F2%5D%3D%201)
Substitute 0 and x/2 for y
Integrate x
![k * \frac{x^2}{2*2} [0,2]= 1](https://tex.z-dn.net/?f=k%20%2A%20%5Cfrac%7Bx%5E2%7D%7B2%2A2%7D%20%5B0%2C2%5D%3D%201)
![k * \frac{x^2}{4} [0,2]= 1](https://tex.z-dn.net/?f=k%20%2A%20%5Cfrac%7Bx%5E2%7D%7B4%7D%20%5B0%2C2%5D%3D%201)
Substitute 0 and 2 for x
![k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%20%5Cfrac%7B2%5E2%7D%7B4%7D%20-%20%5Cfrac%7B0%5E2%7D%7B4%7D%20%5D%3D%201)
![k *[ \frac{4}{4} - \frac{0}{4} ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%20%5Cfrac%7B4%7D%7B4%7D%20-%20%5Cfrac%7B0%7D%7B4%7D%20%5D%3D%201)
![k *[ 1-0 ]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%201-0%20%5D%3D%201)
![k *[ 1]= 1](https://tex.z-dn.net/?f=k%20%2A%5B%201%5D%3D%201)
Solving (b): 
We have:
Where
To find , we use:
So, we have:
Integrate x leave y
![P(x > 3y) = \int\limits^2_0 x [0,y/3]dy](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cint%5Climits%5E2_0%20%20x%20%5B0%2Cy%2F3%5Ddy)
Substitute 0 and y/3 for x
![P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cint%5Climits%5E2_0%20%20%5By%2F3%20-%200%5Ddy)
Integrate
![P(x > 3y) = \frac{y^2}{2*3} [0,2]](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cfrac%7By%5E2%7D%7B2%2A3%7D%20%5B0%2C2%5D)
![P(x > 3y) = \frac{y^2}{6} [0,2]\\](https://tex.z-dn.net/?f=P%28x%20%3E%203y%29%20%3D%20%5Cfrac%7By%5E2%7D%7B6%7D%20%5B0%2C2%5D%5C%5C)
Substitute 0 and 2 for y
