Given:
In January Joanna deposited $250 into her savings account.
In February, she deposited an additional $100.
Her account has an APR of 6% compounded monthly.
Required:
We have to find how much interest did Joanna earn in the first two months.
Step-by-step explanation:
For the month of January:
![A=P(1+(r)/(100))^n](https://img.qammunity.org/2023/formulas/mathematics/high-school/sg4zk7ybldcrek0y9wnq377cpzs7p93hjn.png)
Here, P=$250, r=6%, amd n= 1 month=1/12 year.
Then,
![\begin{gathered} A=250(1+(6)/(100))^{(1)/(12)} \\ \\ A=250((106)/(100))^{(1)/(12)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zy93tgtpt56e7mm0541zesfis7629au9qo.png)
![\begin{gathered} A=250*1.005 \\ A=\text{ \$}251.25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f048bahw43x88zrmhi80pgl9nrfnh6kgpy.png)
Then the interest is
![I=A-P=251.25-250=\text{ \$}1.25](https://img.qammunity.org/2023/formulas/mathematics/college/tu9ejjxyjlyvi2p9ru4d36fxl5iipp3kzh.png)
For the month of February:
P=251.24+100=351.25
Then we have
![\begin{gathered} A=351.25(1+(6)/(100))^{(1)/(12)} \\ \\ A=351.25((106)/(100))^{(1)/(12)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7g1mgfzyxsnyoggrig1f2jxs3nyqxstx2w.png)
![\begin{gathered} A=351.5*1.005 \\ A=\text{ \$}353.01 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/26fp3ry7zxupzv6xc3zhcv4inr0kxmq9uh.png)
Then the interest is
![A=P-I=353.01-351.25=\text{ \$}1.76](https://img.qammunity.org/2023/formulas/mathematics/college/kqfgjwxb3tdod23aobno8hrwe4bmjxnck9.png)
Therefore, the total interest is
![1.25+1.76=\text{ \$}3.01](https://img.qammunity.org/2023/formulas/mathematics/college/iswehgqxybsy9vidhpois47v7mvgvx1omh.png)
Final answer:
Hence the final answer is
![\text{ \$}3.01](https://img.qammunity.org/2023/formulas/mathematics/college/wl7ilcek4zzvus0mk2uaum6zccvrhbuj0y.png)