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

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)

Responses

$30,314.09
$30,314.09

$7,716.96
$7,716.96

$7,600.00
$7,600.00

$8,840.80

Answer 1

To find the value of the investment account after 7 years at an interest rate of 2.85%, we need to substitute the value of x into the equation C(x).

First, we need to calculate the value of x. Since x = 1 + r, and r = 2.85%, we have:
x = 1 + 0.0285
x = 1.0285

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

Using a calculator, we find that C(1.0285) is approximately:
C(1.0285) ≈ $7,716.96

Therefore, the value of the investment account after 7 years at an interest rate of 2.85% is $7,716.96.

Answer 2

To find the value of the investment account after 7 years with an interest rate of 2.85%, we can use the given equation:

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

where x = 1 + r.

Substituting the given interest rate, we have:

x = 1 + 0.0285
x = 1.0285

Now we can plug in this value of x into the equation and calculate C(x):

C(x) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4

C(x) = 1500(1.223864837) + 2000(1.195618027) + 1870(1.166329273) + 2230(1.146454257)

C(x) = 1835.797256 + 2391.236054 + 2181.118987 + 2553.715292

C(x) = 8961.867589

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is $8,961.87.

Answer 3

To find the value of the investment account after 7 years with an interest rate of 2.85%, we can plug in the values into the equation C(x) = 1500x7 + 2000x6 + 1870x5 + 2230x4.

First, let's calculate the value of x. X can be calculated using the formula x = 1 + r, where r is the interest rate.

x = 1 + 2.85% = 1 + 0.0285 = 1.0285

Next, let's calculate the value of C(x).

C(x) = 1500(x^7) + 2000(x^6) + 1870(x^5) + 2230(x^4)
C(x) = 1500(1.0285^7) + 2000(1.0285^6) + 1870(1.0285^5) + 2230(1.0285^4)

Using a calculator, you can evaluate the above equation to find the value of the investment account after 7 years with an interest rate of 2.85%.

Calculating this, we get:

C(x) = 1500(1.0285^7) + 2000(1.0285^6) + 1870(1.0285^5) + 2230(1.0285^4)
C(x) = 1500(1.2048) + 2000(1.1829) + 1870(1.1604) + 2230(1.1381)
C(x) = 1807.2 + 2365.8 + 2169.48 + 2539.43
C(x) = 8876.91

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is $8,876.91.

None of the given options match the calculated value of the investment account after 7 years with an interest rate of 2.85%, which is $8,876.91.