If $11,100 is invested at 3.1% interest compounded monthly, how much will the investment be worth in 19 years? A.$19,914.08 B.$19,826.11 C.$19,958.89 D.$19,989.07
is it B?
It depends if the 3.1% is an annual rate...which I suppose it is. The formula would be:\[A=P(1+\frac{r}{n})^{nt}\] Where P = principal r = annual rate of interest (as a decimal) t = number of years A = amount of money accumulated after n years, including interest. n = number of times the interest is compounded per year \[A=11100(1+\frac{0.031}{12})^{(12)(19)}=?\]
I have made a mistake 3.1% is nominal because of that you need to divide by 12 i = 3.1/12
future value is D
@Shane_B is right!
If you calculate it out, you should end up with answer D
thank you
yw
Beginning balance on the monthly bank statement for Charmaine's checking account was $603.84 and ending balance was $264.98. What can be said about Charmaine's transactions for the month? A. She had $868.82 more in debits than in credits B. she had $868.82 more in credits than in debits C. She had $338.86 more in credits than in debits D. she had $338.86 more in debits than in credits
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