Question

Question 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)

Answer 1

To calculate the value of Brayton's investment account after 7 years, we can use the provided equation:

\[ C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4 \]

where \( x = 1 + r \) and \( r \) is the interest rate. Given an interest rate of \( r = 2.85% \), we first convert it to decimal form:

\[ r = 0.0285 \]

Now, we can find \( x \):

\[ x = 1 + r = 1 + 0.0285 = 1.0285 \]

Next, we will substitute \( x \) into the equation \( C(x) \):

\[ C(1.0285) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4 \]

Now, let's calculate each term step by step:

1. Calculate \( (1.0285)^4 \): \[ (1.0285)^4 \approx 1.1171 \text{ (approximately)} \]

2. Calculate \( (1.0285)^5 \): \[ (1.0285)^5 \approx 1.1496 \text{ (approximately)} \]

3. Calculate \( (1.0285)^6 \): \[ (1.0285)^6 \approx 1.1830 \text{ (approximately)} \]

4. Calculate \( (1.0285)^7 \): \[ (1.0285)^7 \approx 1.2176 \text{ (approximately)} \]

Now we can substitute these values back into the equation:

\[ C(1.0285) \approx 1500(1.2176) + 2000(1.1830) + 1870(1.1496) + 2230(1.1171) \]

Calculating each term:

1. \( 1500 \times 1.2176 \approx 1826.4 \)
2. \( 2000 \times 1.1830 \approx 2366.0 \)
3. \( 1870 \times 1.1496 \approx 2146.76 \)
4. \( 2230 \times 1.1171 \approx 2492.57 \)

Now sum these amounts:

\[ C(1.0285) \approx 1826.4 + 2366.0 + 2146.76 + 2492.57 \]

Calculating the total:

\[ C(1.0285) \approx 1826.4 + 2366.0 = 4192.4 \] \[ 4192.4 + 2146.76 \approx 6339.16 \] \[ 6339.16 + 2492.57 \approx 8831.73 \]

Thus, the value of the investment account after 7 years at an interest rate of 2.85% is approximately:

\[ \boxed{8831.73} \]

Answer 2

Not a choice

Answer 3

Let's go through the calculations again step-by-step to ensure accuracy.

1. Calculate \( x \): \[ x = 1 + r = 1 + 0.0285 = 1.0285 \]

2. Calculate powers of \( x \):

   • \( x^4 \): \[ (1.0285)^4 \approx 1.117789 \]
   • \( x^5 \): \[ (1.0285)^5 \approx 1.149574 \]
   • \( x^6 \): \[ (1.0285)^6 \approx 1.18314 \]
   • \( x^7 \): \[ (1.0285)^7 \approx 1.217697 \]

3. Plug these values back into C(x): \[ C(1.0285) = 1500(1.217697) + 2000(1.18314) + 1870(1.149574) + 2230(1.117789) \]

   Now, calculate each term:

   • First Term: \[ 1500 \times 1.217697 \approx 1826.5455 \]
   • Second Term: \[ 2000 \times 1.18314 \approx 2366.28 \]
   • Third Term: \[ 1870 \times 1.149574 \approx 2146.61 \]
   • Fourth Term: \[ 2230 \times 1.117789 \approx 2492.59 \]

4. Sum these values: \[ C(1.0285) \approx 1826.5455 + 2366.28 + 2146.61 + 2492.59 \]

   • First add the first two: \[ 1826.5455 + 2366.28 \approx 4192.8255 \]
   • Next, add the third: \[ 4192.8255 + 2146.61 \approx 6339.4355 \]
   • Finally, add the fourth: \[ 6339.4355 + 2492.59 \approx 8832.0255 \]

Thus, the final value of the investment account after 7 years at an interest rate of 2.85% is approximately:

\[ \boxed{8832.03} \]

If this is still not matching the choices provided, please check the options or any conditions in the problem that might have been missed.