Answer:
I'm not an expert here, this is a best guess!
But I would say if there is no chance that of him incurring excess costs of less than $500, then he knows without insurance he'll end up paying at least $500, possibly more out of pocket, without the insurance.
so I would say He ends up spending the least amount out if pocket by going with option A. for $75. that's $75 out of pocket with no deductible and it covers his $500+ in excess costs....B and C would also cover the excess, but would each cost $140 or $275 out of pocket at the end of the day....
with that being said, I'd say it's worth it to buy the insurance....even if he doesn't have any excess costs, he's spent $75 dollars for the peace of mind to know he's covered either way, and if he does incur the excess costs he's spent $75 rather that $500+....Even if the excess charges are only $100, which it says there is no chance of happening, but still, then he's still saved $25 altogether. Unless I'm reading it wrong, Option A saves him the most money either way, and is worth it to buy the insurance!
volume of the ball = 4/3 * PI *r^3 =
4/3 * 3.14 * 2.86^3 = 97.99 cubic inches
this is less than the volume of the box, so yes it will fit
Answer:
a) 
And replacing we got:
b) 
And then the expected value would be:
Step-by-step explanation:
We assume the following distribution given:
Y 0 1 2 3
P(Y) 0.60 0.25 0.10 0.05
Part a
We can find the expected value with this formula:
And replacing we got:
Part b
If we want to find the expected value of 
we need to find the expected value of Y^2 and we have:
And replacing we got:
And then the expected value would be:
Answer:
28
Step-by-step explanation:
f(6) = -3 * 6 + 10 = -18 + 10 = -8
g(f(6)) = g(-8)
g(-8) = 
= 64 - 24 - 12 = 64 - 36 = 28
