Round 363 to the nearest tens and hundreds
1 answer:
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Answer:
We want to find:
Here we can use Stirling's approximation, which says that for large values of n, we get:
Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
Now we can just simplify this, so we get:
And we can rewrite it as:
The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus: