Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
To determine which exponential function models the data given in the table, let's examine the account values over the time period and compare them to the provided exponential function candidates.
We are given the following data points for the account value over time:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (years)} & \text{Account Value (\$)} \\ \hline 0 & 400 \\ \hline 1 & 436 \\ \hline 2 & 468 \\ \hline 3 & 502 \\ \hline 4 & 550 \\ \hline 5 & 589 \\ \hline \end{array} \][/tex]
We'll compare these data points to each candidate exponential function to see which one fits:
1. [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
2. [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
3. [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
4. [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]
For each candidate function, we substitute [tex]\( x \)[/tex] with the time values from 0 to 5 and compare the resulting values to the account values given.
Let's check if these functions match the data points:
- [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
- [tex]\( f(0) = 1.41 \cdot (30.69)^0 = 1.41 \)[/tex]
- This does not match the initial account value of 400.
- [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
- [tex]\( f(0) = 37.74 \cdot (396.48)^0 = 37.74 \)[/tex]
- This does not match the initial account value of 400.
- [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
- [tex]\( f(0) = 390.60 \cdot (1.41)^0 = 390.60 \)[/tex]
- This is close but does not exactly match the initial account value of 400. Checking further might show divergences.
- [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]
- [tex]\( f(0) = 401.19 \cdot (1.08)^0 = 401.19 \)[/tex]
- This closely matches the initial account value of 400. Checking more values:
- [tex]\( f(1) = 401.19 \cdot (1.08)^1 = 401.19 \cdot 1.08 \approx 433.29 \)[/tex]
- [tex]\( f(2) = 401.19 \cdot (1.08)^2 \approx 467.95 \)[/tex]
- [tex]\( f(3) = 401.19 \cdot (1.08)^3 \approx 504.39 \)[/tex]
- [tex]\( f(4) = 401.19 \cdot (1.08)^4 \approx 543.86 \)[/tex]
- [tex]\( f(5) = 401.19 \cdot (1.08)^5 \approx 587.37 \)[/tex]
These values closely match the provided account values when rounded to the nearest hundredth and are closer than any other function candidate. Thus, the exponential function that models the data is:
[tex]\[ f(x) = 401.19 \cdot (1.08)^x \][/tex]
We are given the following data points for the account value over time:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (years)} & \text{Account Value (\$)} \\ \hline 0 & 400 \\ \hline 1 & 436 \\ \hline 2 & 468 \\ \hline 3 & 502 \\ \hline 4 & 550 \\ \hline 5 & 589 \\ \hline \end{array} \][/tex]
We'll compare these data points to each candidate exponential function to see which one fits:
1. [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
2. [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
3. [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
4. [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]
For each candidate function, we substitute [tex]\( x \)[/tex] with the time values from 0 to 5 and compare the resulting values to the account values given.
Let's check if these functions match the data points:
- [tex]\( f(x) = 1.41 \cdot (30.69)^x \)[/tex]
- [tex]\( f(0) = 1.41 \cdot (30.69)^0 = 1.41 \)[/tex]
- This does not match the initial account value of 400.
- [tex]\( f(x) = 37.74 \cdot (396.48)^x \)[/tex]
- [tex]\( f(0) = 37.74 \cdot (396.48)^0 = 37.74 \)[/tex]
- This does not match the initial account value of 400.
- [tex]\( f(x) = 390.60 \cdot (1.41)^x \)[/tex]
- [tex]\( f(0) = 390.60 \cdot (1.41)^0 = 390.60 \)[/tex]
- This is close but does not exactly match the initial account value of 400. Checking further might show divergences.
- [tex]\( f(x) = 401.19 \cdot (1.08)^x \)[/tex]
- [tex]\( f(0) = 401.19 \cdot (1.08)^0 = 401.19 \)[/tex]
- This closely matches the initial account value of 400. Checking more values:
- [tex]\( f(1) = 401.19 \cdot (1.08)^1 = 401.19 \cdot 1.08 \approx 433.29 \)[/tex]
- [tex]\( f(2) = 401.19 \cdot (1.08)^2 \approx 467.95 \)[/tex]
- [tex]\( f(3) = 401.19 \cdot (1.08)^3 \approx 504.39 \)[/tex]
- [tex]\( f(4) = 401.19 \cdot (1.08)^4 \approx 543.86 \)[/tex]
- [tex]\( f(5) = 401.19 \cdot (1.08)^5 \approx 587.37 \)[/tex]
These values closely match the provided account values when rounded to the nearest hundredth and are closer than any other function candidate. Thus, the exponential function that models the data is:
[tex]\[ f(x) = 401.19 \cdot (1.08)^x \][/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
