Brayton wants to invest his high school earnings for the next 7 years. He deposits $1,500 into an investment account at the beginning of the first year, $2,000 at the beginning of the second year, $1,870 at the beginning of the third year, and $2,230 at the beginning of the fourth year. Use x=1+r , where r is the interest rate, and the equation C(x)=1500x7+2000x6+1870x5+2230x4 . The equation represents the relationship between C(x), the value of the investment after 7 years, and its annual interest rate, r. Find the value of the investment account if the interest rate is 2.85%.(1 point)

Question

Responses

$30,314.09
$30,314.09

$8,840.80
$8,840.80

$7,716.96
$7,716.96

$7,600.00

Answer 1

To find the value of the investment account after 7 years using the formula \( C(x) \) provided, we first need to substitute the interest rate into the equation.

Given the interest rate \( r = 2.85% \), we convert it to decimal form: \[ r = 0.0285 \]

Now, we calculate \( x \): \[ x = 1 + r = 1 + 0.0285 = 1.0285 \]

Next, we substitute \( x = 1.0285 \) into the equation \( C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4 \):

1. Calculate \( x^7 \): \[ x^7 = (1.0285)^7 \approx 1.211156 \]

2. Calculate \( x^6 \): \[ x^6 = (1.0285)^6 \approx 1.177864 \]

3. Calculate \( x^5 \): \[ x^5 = (1.0285)^5 \approx 1.146163 \]

4. Calculate \( x^4 \): \[ x^4 = (1.0285)^4 \approx 1.116348 \]

Now we can substitute back into the equation: \[ C(x) = 1500(1.211156) + 2000(1.177864) + 1870(1.146163) + 2230(1.116348) \]

Calculating each term:

1. \( 1500 \times 1.211156 \approx 1816.74 \)
2. \( 2000 \times 1.177864 \approx 2355.73 \)
3. \( 1870 \times 1.146163 \approx 2143.09 \)
4. \( 2230 \times 1.116348 \approx 2492.26 \)

Now add these values together: \[ C(x) \approx 1816.74 + 2355.73 + 2143.09 + 2492.26 \approx 8807.82 \]

Thus, the value of the investment account if the interest rate is 2.85% is approximately $8,807.82.

Looking at the choices provided, the closest answer based on the calculations would be $8,840.80.