Answer:
The final cost is 28249*(1+3.49/100)^5=33534.74
Answer:
- <u>59.0891 g (rounded to 4 decimal places)</u>
Explanation:
<em>Half-life time</em> of a radioactive substance is the time for half of the substance to decay.
Thus, the amount of the radioactive substance that remains after a number n of half-lives is given by:
Where:
- A is the amount that remains of the substance after n half-lives have elapses, and
- A₀ is the starting amount of the substance.
In this problem, you have that the half-live for your sample (polonium-210) is 138 days and the number of days elapsed is 330 days. Thus, the number of half-lives elapsed is:
- 330 days / 138 days = 2.3913
Therefore, the amount of polonium-210 that will be left in 330 days is:
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!
Answer:
a) uniform
b) 1/2
c) 1/1000
Step-by-step explanation:
a) "numbers with equal probability" have a uniform distribution.
__
b) Even numbers make up 1/2 of all numbers.
__
c) There are ten such numbers in the range, so the probability is ...
10/10000 = 1/1000
