Madeline invested $$ \$ 2,800 $$ in an account paying an interest rate of $$ 4.4 \% $$ compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $$ \$ 6,100 $$ ?

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To determine how long it will take for Madeline's investment to grow from \$2,800 to \$6,100 with an annual compound interest rate of 4.4%, we can use the **compound interest formula**:

$$
A = P \times (1 + r)^t
$$

Where:
- $$ A $$ = the amount of money accumulated after $$ t $$ years (\$6,100)
- $$ P $$ = the principal investment amount (\$2,800)
- $$ r $$ = the annual interest rate (4.4% or 0.044)
- $$ t $$ = the number of years

**Step 1: Set Up the Equation**

$$
6,100 = 2,800 \times (1 + 0.044)^t
$$

**Step 2: Divide Both Sides by 2,800**

$$
\frac{6,100}{2,800} = (1.044)^t
$$

$$
2.178571429 \approx (1.044)^t
$$

**Step 3: Apply Natural Logarithm to Both Sides**

$$
\ln(2.178571429) = t \times \ln(1.044)
$$

**Step 4: Calculate the Natural Logarithms**

$$
\ln(2.178571429) \approx 0.7793
$$
$$
\ln(1.044) \approx 0.0430
$$

**Step 5: Solve for $$ t $$**

$$
t = \frac{0.7793}{0.0430} \approx 18.1 \text{ years}
$$

**Conclusion:**

It will take approximately **18.1 years** for Madeline's investment to grow to \$6,100.

**Final Answer:**  
Approximately 18.1 years are needed for the account to reach $6,100.
It will take approximately 18.1 years for Madeline's investment to reach \$6,100.

Madeline invested in an account paying an interest rate of compounded annually. Assuming no
deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account
to reach ?

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Madeline invested in an account paying an interest rate of compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach ?