Question

v(n) | 2200 | 2287.96 | 2527.11 | 2873.84 | 2902.85 | 3044.74
use your calculator to determine the exponential regression equation that models the set of data above. round the \a\ value to two decimals, and round the \b\ value to three decimals. use the indicated variables and proper function notation.
v(n) = 2208.06(1.019)^n
based on the your regression model, what is the percent increase per year?
2
find v(13). round your answer to two decimal places.
v(13) = 2802.62
interpret your answer by completing the following sentence. be sure to use units in your answer.
\the value of the investment after 13 years is $2803 .\
how long will it take for the value of this investment to reach $10,000? round your answer to two decimal places.
v(n) = 10000 when n = 81
interpret your answer by completing the following sentence. be sure to use units in your answer.
in 81.02 , the value of the investment will reach $10,000 .
how long will it take for the value of the investment to double? round your answer to two decimal places.
v(n) has doubled when n = 37.05
interpret your answer by completing the following sentence. be sure to use units in your answer.
in 37.05 , the value of the investment will double.

Explanation:

Step1: Percent Increase

The exponential regression model is \( V(n) = 2208.06(1.019)^n \). The general form of exponential growth is \( y = a(b)^n \), where \( b = 1 + r \), and \( r \) is the growth rate. So, \( 1 + r = 1.019 \), so \( r = 1.019 - 1 = 0.019 \), which is \( 1.9\% \).

Step2: Find \( V(13) \)

Substitute \( n = 13 \) into \( V(n) = 2208.06(1.019)^{13} \). Calculate \( (1.019)^{13} \approx 1.270 \) (using calculator). Then \( V(13) \approx 2208.06 \times 1.270 \approx 2804.24 \).

Step3: Time to Reach $10,000

Set \( 2208.06(1.019)^n = 10000 \). Divide both sides by 2208.06: \( (1.019)^n = \frac{10000}{2208.06} \approx 4.529 \). Take natural log: \( n \ln(1.019) = \ln(4.529) \). So \( n = \frac{\ln(4.529)}{\ln(1.019)} \approx \frac{1.510}{0.0188} \approx 80.32 \).

Step4: Time to Double

Set \( 2208.06(1.019)^n = 2 \times 2208.06 \), so \( (1.019)^n = 2 \). Take natural log: \( n \ln(1.019) = \ln(2) \). So \( n = \frac{\ln(2)}{\ln(1.019)} \approx \frac{0.6931}{0.0188} \approx 36.87 \).

Answer:

• Percent Increase: \( 1.9\% \)
• \( V(13) \approx 2804.24 \)
• Time to $10,000: \( \approx 80.32 \) years
• Time to Double: \( \approx 36.87 \) years