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

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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:
$$
\boxed{694.51}
$$
The total balance after 2 years is approximately \$694.51.

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