To model Bertha's investment account using the given deposits and an interest rate, we first need to recognize how each deposit compounds over time.
Deposits and their compounding:
- The first deposit of $1,200 is made at the beginning of year 1, so it will be in the account for 3 years.
- The second deposit of $880 is made at the beginning of year 2, so it will be in the account for 2 years.
- The third deposit of $830 is made at the beginning of year 3, so it will be in the account for 1 year.
Let's combine these into a polynomial model \( C(x) \) where \( x = 1 + r \).
Using \( r = 0.056 \) (5.6 percent), we can calculate \( x \): \[ x = 1 + r = 1 + 0.056 = 1.056 \]
Construct the polynomial: The amount of each deposit at the end of the 3 years is as follows:
- For the first deposit of $1,200: \[ 1200 \cdot x^3 \]
- For the second deposit of $880: \[ 880 \cdot x^2 \]
- For the third deposit of $830: \[ 830 \cdot x^1 \]
Combining these, we have the polynomial: \[ C(x) = 1200x^3 + 880x^2 + 830x \]
Now plug in \( x = 1.056 \) and calculate \( C(1.056) \):
- Calculate \( x^3 \): \[ (1.056)^3 \approx 1.191016 \]
- Calculate \( x^2 \): \[ (1.056)^2 \approx 1.115936 \]
- Calculate \( x \): \[ (1.056)^1 = 1.056 \]
Now substituting back into the polynomial: \[ C(1.056) = 1200 \cdot 1.191016 + 880 \cdot 1.115936 + 830 \cdot 1.056 \] Calculating each component:
- \( 1200 \cdot 1.191016 \approx 1429.2192 \)
- \( 880 \cdot 1.115936 \approx 981.9128 \)
- \( 830 \cdot 1.056 \approx 875.88 \)
Finally, sum these amounts: \[ C(1.056) \approx 1429.2192 + 981.9128 + 875.88 \approx 3287.012 \]
Final Calculation: Rounding this to two decimal places, the final amount of Bertha’s investment account after 3 years is approximately: \[ \boxed{3287.01} \]