The value of can never be 0
y = 0
A
%error=(6.5g/mol - 7g/mol)/(7g/mol) * 100= 7.14% error
Answer:
it should be A but im not fully sure
Answer: -1
Step-by-step explanation: f(x) is another way of saying y. So when y is 3, according to the table, x is -1
Answer:
1) 
2) 
3) 
And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
4) 
And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be:
Step-by-step explanation:
For this case we have the following distributions given:
Probability M J
0.3 14% 22%
0.4 10% 4%
0.3 19% 12%
Part 1
The expected value is given by this formula:

And replacing we got:

Part 2

Part 3
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89](https://tex.z-dn.net/?f=Var%20%28M%29%3D%20E%28M%5E2%29%20-%5BE%28M%29%5D%5E2%20%3D%20207.1%20-%2813.9%5E2%29%3D%2013.89)
And the deviation would be:
Part 4
We can calculate the second moment first with the following formula:

And the variance would be given by:
![Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56](https://tex.z-dn.net/?f=Var%20%28J%29%3D%20E%28J%5E2%29%20-%5BE%28J%29%5D%5E2%20%3D%20194.8%20-%2811.8%5E2%29%3D%2055.56)
And the deviation would be: