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) = 1500x ^ 7 + 2000x ^ 6 + 1870x ^ 5 + 2230x ^ 4 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) < $7,716.96 $8,840.80 $30,314.09 $7,600.00

Answer 1

To find the value of the investment account after 7 years with an interest rate of 2.85%, we first need to calculate $x$ using the formula $x=1+r$:

$$r=0.0285\Rightarrow x=1+0.0285=1.0285$$

Next, we need to plug this value of $x$ into the equation for $C(x)$:

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

Now, we will calculate each term separately:

1. $1500x^7=1500({1.0285}^7)$
2. $2000x^6=2000({1.0285}^6)$
3. $1870x^5=1870({1.0285}^5)$
4. $2230x^4=2230({1.0285}^4)$

Now let's calculate $x^4$, $x^5$, $x^6$, and $x^7$:

• ${1.0285}^4\approx 1.1171$
• ${1.0285}^5\approx 1.1485$
• ${1.0285}^6\approx 1.1803$
• ${1.0285}^7\approx 1.2127$

Next, we calculate each term:

1. $1500(1.2127)\approx 1819.05$
2. $2000(1.1803)\approx 2360.60$
3. $1870(1.1485)\approx 2147.85$
4. $2230(1.1171)\approx 2495.69$

Now, let's sum these values to find $C(x)$:

$$C(x)=1819.05+2360.60+2147.85+2495.69=8853.19$$

The value of the investment account after 7 years at an interest rate of 2.85% is approximately $8,853.19.

From the provided options, it appears that the closest choice is $8,840.80.