Advance Algebra Logarithm Task
1 answer:
9514 1404 393
Answer:
3.65% monthly
Step-by-step explanation:
The same amount is invested for the same period in all accounts, so we only need to determine the effective annual rate in order to compare the accounts.
For compounding annual rate r n times per year, the effective annual rate is ...
(1 +r/n)^n -1
For the same rate r, larger values of n cause effective rate to be higher. As a consequence, we know that 3.65% compounded quarterly will not have as great a yield as 3.65% compounded monthly. The effective rate for the monthly compounding is ...
(1 +0.0365/12)^12 -1 = 3.712%
The effective rate for continuous compounding is ...
e^r -1
For a continuously compounded rate of 3.6%, the effective annual rate is ...
e^0.036 -1 = 3.666%
This tells us the best yield is in the account bearing 3.65% compounded monthly.
_____
If i is the effective annual rate of interest as computed by the methods above, then the 10-year account balance will be ...
10000×(1 +i)^10
This is the formula used in the spreadsheet to calculate the balances shown.
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Answer:
After 9 years, the population reach 4000.
Step-by-step explanation:
The given function P(t) represents the size of a small herbivore population at time t (in years).
We need to find the number of years after which population reach 4000.
Substitute P(t)=4000 in the above function.
Divide both sides by 1000.
Taking ln on both sides.
Divide both sides by 0.16.
We need to find the number of years. So, round the answer to the next whole number.
Therefore, after 9 years, the population reach 4000.