Solution:
The image shows a mathematics word problem about savings account interest. Unfortunately, the image is quite blurry, which makes it difficult to discern all of the details with precision. However, I can gather the general information that:- There is a savings account with an annual percentage rate (interest rate) of 3.6%.- The monthly deposit is $200 into the account.- You are asked to calculate the amount of money that will be in the account in 1 year.To solve it, assuming the interest compounds monthly, we will use the formula for the future value of a series of equal payments made at regular intervals, compounded at a regular frequency:\[ A = P \frac{(1 + r/n)^{nt} - 1}{r/n} \]Where:- $$ A $$ is the future value of the investment/loan, including interest- $$ P $$ is the payment amount per period- $$ r $$ is the annual interest rate (decimal)- $$ n $$ is the number of times that interest is compounded per year- $$ t $$ is the time the money is invested or borrowed for, in yearsFor this question:- $$ P = $200 $$- $$ r = 3.6\% $$ or 0.036 as a decimal- $$ n = 12 $$ (since it compounds monthly)- $$ t = 1 $$ yearUsing the formula:\[ A = 200 \frac{(1 + 0.036/12)^{12*1} - 1}{0.036/12} \]First calculate $$ 0.036/12 $$ and add 1 to it:\[ 1 + 0.036/12 = 1.003 \]Now raise this to the power of $$ 12 * 1 $$, which is 12:\[ (1.003)^{12} \approx 1.0367 \]Now compute $$ (1.003)^{12} - 1 $$:\[ 1.0367 - 1 = 0.0367 \]Now divide this result by $$ 0.036/12 $$:\[ A = 200 \times \frac{0.0367}{0.003} \]Perform the division:\[ A = 200 \times 12.2333... \]Now multiply by 200:\[ A \approx 200 \times 12.2333 \approx 2446.666 \]To the nearest penny, this would be rounded to:\[ A \approx \$2446.67 \]So, after 1 year, there would be approximately \$2446.67 in the account. Please note this calculation assumes that interest is compounded monthly and that deposits happen at the end of each month. If any of this information is incorrect due to the lack of clarity in the image, the final calculation may be different.