To model the final amount of Bertha's investment account, we will consider each of her deposits and how they grow over the years with the interest applied.
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Deposit of $1,200 at the beginning of Year 1: This amount will earn interest for 3 years. Hence, the contribution to the final amount will be: \[ 1200 \cdot x^3 \]
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Deposit of $880 at the beginning of Year 2: This amount will earn interest for 2 years. Thus, the contribution to the final amount will be: \[ 880 \cdot x^2 \]
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Deposit of $830 at the beginning of Year 3: This amount will earn interest for 1 year. So, the contribution to the final amount will be: \[ 830 \cdot x^1 \]
Now, we can combine these contributions into the polynomial expressing the total amount in the investment account:
\[ C(x) = 1200x^3 + 880x^2 + 830x \]
Next, we need to find the value of \( C(x) \) when \( x = 1 + r \), with an interest rate \( r = 5.6% \) or \( 0.056\). Therefore:
\[ x = 1 + 0.056 = 1.056 \]
Now we can substitute \( x = 1.056 \) into the polynomial \( C(x) \):
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Calculate \( 1200x^3 \): \[ 1200 \cdot (1.056)^3 \approx 1200 \cdot 1.191016 \approx 1429.22 \]
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Calculate \( 880x^2 \): \[ 880 \cdot (1.056)^2 \approx 880 \cdot 1.118336 \approx 984.24 \]
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Calculate \( 830x \): \[ 830 \cdot (1.056) \approx 830 \cdot 1.056 \approx 876.48 \]
Finally, we will sum these contributions to find the total:
\[ C(1.056) \approx 1429.22 + 984.24 + 876.48 = 3289.94 \]
Thus, the final amount of Bertha’s investment account with an interest rate of 5.6% is approximately $3289.94.
Final answer:
\( C(x) = 1200x^3 + 880x^2 + 830x \)
The final amount of Bertha’s investment account if the interest rate is 5.6 percent is $3289.94.