June has a savings account with an annual simple interest rate of 2.6%. She hopes to gain $6,500 in interest over a period of el

Question

3 years ago

5

June has a savings account with an annual simple interest rate of 2.6%. She hopes to gain $6,500 in interest over a period of el

even years. To accomplish this, June invested $17,801 in the account, but later realized that this was not enough money. To the nearest dollar, how much more money should June have initially invested to reach her goal? a. $6,335 b. $3,715 c. $1,409 d. $4,926

Mathematics

2 answers:

Komok [63] · Answer 1 · 2022-03-06T03:14:40+03:00

Komok [63]3 years ago

8 0

P=6500/(0.026*11)
P=22,727.27 to be invested
So how much more
22727.27-17801=4926.27

leonid [27] · Answer 2 · 2022-03-06T02:00:27+03:00

leonid [27]3 years ago

7 0

$\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$6500\\
P=\textit{original amount deposited}\\
r=rate\to 2.6\%\to \frac{2.6}{100}\to &0.026\\
t=years\to &3
\end{cases}$

she should have invested P, in order to get I = 6500

but she instead invested 17,801, how much more did she need? well, P - 17801

solve for P