Answer:


![V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%3D349.2-%2818.6%29%5E2%3D3.24)
The expected price paid by the next customer to buy a freezer is $466
Step-by-step explanation:
From the information given we know the probability mass function (pmf) of random variable X.

<em>Point a:</em>
- The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

Therefore

- If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29)
So
and
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29%5C%5CE%5BX%5E2%5D%3D%24%5Csum_%7BD%7Dx%5E2%5Ccdot%20p%28x%29%5C%5C%20E%28X%5E2%29%3D16%5E2%5Ccdot%200.3%2B18%5E2%5Ccdot%200.1%2B20%5E2%5Ccdot%200.6%5C%5CE%28X%5E2%29%3D349.2)
- The variance of X, denoted by V(X), is
![V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2](https://tex.z-dn.net/?f=V%28X%29%20%3D%20%24%5Csum_%7BD%7DE%5B%28X-%5Cmu%29%5E2%5D%3DE%28X%5E2%29-%5BE%28X%29%5D%5E2)
Therefore
![V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%5C%5CV%28X%29%3D349.2-%2818.6%29%5E2%5C%5CV%28X%29%3D3.24)
<em>Point b:</em>
We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:
From the rules of expected value this proposition is true:
We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore
The expected price paid by the next customer is

Answer:

Step-by-step explanation:
Fraction of the total that is for corn (TC - Total corn):

fraction of the corn section that is for white corn (WC - white corn in the corn seccion):

we need to find the fraction of the whole field that is for the white corn.
For this we need to find how much is
out of the
destinated to corn, and this will be the fraction of the total that is for white corn. We find this fraction by multiplying the fraction of corn (
) by the fraction of white corn in the corn section (
).
I will call the total fraction of white corn TWC, thus:

the answer is:
of the whole field is planted with white corn
Let ordered pairs
X Y
-1 1
-2 2
-3 3
-4 4
-5 5
<span> What kind of figure do you get?
For this case a straight line was obtained, whose equation is
</span><span> y = -x
</span><span> In what quadrants does the figure lie?
The figure is in the second quadrant. (See graphic)
</span>
Answer:
it should be 187.6 if u got it wrong let me know so i can give u another answer
Step-by-step explanation:
If we let x as candy A
y as candy B
a as dark chocolate in candy a
b as dark chocolate in candy b
c as caramel
d as walnut
P as profit
we have the equations:
a + c = x
2b + d = y
a + 2b ≤ 360
c ≤ 430
d ≤ 210
P = 285x + 260y
This is an optimization problem which involves linear programming. It can be solved by graphical method or by algebraic solution.
P = 285(a + c) + 260(2b +d)
If we assume a = b
Then a = 120, 2b = 240
P = 285(120 + 120) + 260(240 + 120)
P = 162000
candy A should be = 240
candy B should be = 360