Rebs has a savings account earning $$ 1.4 \% $$ simple interest. She has $$ \$ 803 $$ in the account now. How much did she have in the account when she opened it 19 months ago? Round to the nearest cent.

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We start by letting $$ P $$ be the amount in the account when Rebs opened it.

Since the account earns simple interest at an annual rate of $$ 1.4\% $$, the interest earned over a time period $$ t $$ (in years) is

$$
\text{Interest} = P \times 0.014 \times t.
$$

Rebs has had the account for 19 months. Converting months into years, we have

$$
t = \frac{19}{12} \text{ years}.
$$

Thus, the total amount in the account after $$ t $$ years is

$$
P + P \times 0.014 \times \frac{19}{12} = P\left(1 + 0.014 \times \frac{19}{12}\right).
$$

We are told this amount is $$ \$803 $$, so

$$
P\left(1 + 0.014 \times \frac{19}{12}\right) = 803.
$$

First, simplify the factor:

$$
0.014 \times \frac{19}{12} = \frac{0.014 \times 19}{12} = \frac{0.266}{12} \approx 0.02216667.
$$

Then

$$
P(1 + 0.02216667) = P(1.02216667) = 803.
$$

Solving for $$ P $$, we have

$$
P = \frac{803}{1.02216667}.
$$

Calculating the division gives

$$
P \approx 785.59.
$$

Thus, when Rebs opened the account, she had approximately $$ \$785.59 $$.
Rebs had approximately \$785.59 in her account when she opened it 19 months ago.

Rebs has a savings account earning simple interest. She has in the account now. How much did
she have in the account when she opened it 19 months ago? Round to the nearest cent.

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Rebs has a savings account earning simple interest. She has in the account now. How much did she have in the account when she opened it 19 months ago? Round to the nearest cent.