Multiple Cholce 1 point Nicolas has \$650 to deposit into two different savings accounts. - Nicolas will deposit \$400 into Account I, which earns \( 3.5 \% \) annual simple interest. - He will deposit \( \$ 250 \) into Account II, which earns y \& \( \% \) interest compounded annually, Nicolas will not make any additional deposits or withdrawals. Which amount is closest to the total balance of these two accounts at the end of 2 years? \( \$ 672.13 \) \( \$ 694.51 \) \( \$ 694.25 \) \( \$ 695.00 \) Clear my selection

To find the total balance of Nicolas's two accounts at the end of 2 years, we will calculate the balance for each account separately and then sum them up. ### Step 1: Calculate the balance for Account I Account I has a deposit of \$400 with an annual simple interest rate of \(3.5\%\). The formula for calculating the total amount \(A\) with simple interest is: \[ A = P(1 + rt) \] where: - \(P\) = principal amount (initial deposit) - \(r\) = annual interest rate (as a decimal) - \(t\) = time in years For Account I: - \(P = 400\) - \(r = 0.035\) - \(t = 2\) Substituting these values into the formula: \[ A_1 = 400(1 + 0.035 \cdot 2) \] ### Step 2: Calculate the balance for Account II Account II has a deposit of \$250 with an interest rate of \(y\%\) compounded annually. The formula for calculating the total amount \(A\) with compound interest is: \[ A = P(1 + r)^t \] where: - \(P\) = principal amount (initial deposit) - \(r\) = annual interest rate (as a decimal) - \(t\) = time in years For Account II: - \(P = 250\) - \(r = \frac{y}{100}\) - \(t = 2\) Substituting these values into the formula: \[ A_2 = 250\left(1 + \frac{y}{100}\right)^2 \] ### Step 3: Total balance after 2 years The total balance \(A_{total}\) after 2 years is: \[ A_{total} = A_1 + A_2 \] Now, we will calculate \(A_1\) and \(A_2\) and find \(A_{total}\). However, we need to know the value of \(y\) to proceed with the calculation for Account II. Since the problem does not provide \(y\), we will assume \(y\) is such that the total balance is one of the provided options. Let's calculate \(A_1\) first: \[ A_1 = 400(1 + 0.035 \cdot 2) = 400(1 + 0.07) = 400 \cdot 1.07 = 428 \] Now, we will calculate \(A_{total}\) for each option to find the corresponding \(y\). ### Step 4: Calculate \(A_{total}\) for each option 1. For \( \$672.13 \): \[ A_2 = 672.13 - 428 = 244.13 \] \[ 250\left(1 + \frac{y}{100}\right)^2 = 244.13 \] \[ \left(1 + \frac{y}{100}\right)^2 = \frac{244.13}{250} \approx 0.97652 \] \[ 1 + \frac{y}{100} \approx \sqrt{0.97652} \approx 0.9882 \] \[ \frac{y}{100} \approx -0.0118 \Rightarrow y \approx -1.18\% \text{ (not valid)} \] 2. For \( \$694.51 \): \[ A_2 = 694.51 - 428 = 266.51 \] \[ 250\left(1 + \frac{y}{100}\right)^2 = 266.51 \] \[ \left(1 + \frac{y}{100}\right)^2 = \frac{266.51}{250} \approx 1.06604 \] \[ 1 + \frac{y}{100} \approx \sqrt{1.06604} \approx 1.0325 \] \[ \frac{y}{100} \approx 0.0325 \Rightarrow y \approx 3.25\% \] 3. For \( \$694.25 \): \[ A_2 = 694.25 - 428 = 266.25 \] \[ 250\left(1 + \frac{y}{100}\right)^2 = 266.25 \] \[ \left(1 + \frac{y}{100}\right)^2 = \frac{266.25}{250} \approx 1.065 \] \[ 1 + \frac{y}{100} \approx \sqrt{1.065} \approx 1.032 \] \[ \frac{y}{100} \approx 0.032 \Rightarrow y \approx 3.2\% \] 4. For \( \$695.00 \): \[ A_2 = 695.00 - 428 = 267.00 \] \[ 250\left(1 + \frac{y}{100}\right)^2 = 267.00 \] \[ \left(1 + \frac{y}{100}\right)^2 = \frac{267.00}{250} \approx 1.068 \] \[ 1 + \frac{y}{100} \approx \sqrt{1.068} \approx 1.033 \] \[ \frac{y}{100} \approx 0.033 \Rightarrow y \approx 3.3\% \] ### Conclusion The closest total balance of the two accounts at the end of 2 years is \( \$694.51 \). Thus, the answer is: \[ \boxed{694.51} \]