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e value of a certain investment over time is given in the table below. Answer the questions below to determine what kind of fun...Question
e value of a certain investment over time is given in the table below. Answer the questions below to determine what kind of function would best fit the data, linear or exponential.
Number of Years Since Investment Made, x 1 2 3 4
Value of Investment ($), f(x) 13,624.04 11,149.46 9,106.06 7,496.90
Answer
_______ function would best fit the data because as xx increases, the yy values change
_______. The
_______ of this function is approximately.
A linear or exponential
Additively or multiplicatively
Common difference/slope or common ratio/multiplier/base
Number of Years Since Investment Made, x 1 2 3 4
Value of Investment ($), f(x) 13,624.04 11,149.46 9,106.06 7,496.90
Answer
_______ function would best fit the data because as xx increases, the yy values change
_______. The
_______ of this function is approximately.
A linear or exponential
Additively or multiplicatively
Common difference/slope or common ratio/multiplier/base
Answers
GPT-4o mini
To determine what kind of function best fits the given investment data, let's analyze the provided values:
| Number of Years Since Investment Made, x | Value of Investment ($), f(x) |
|-------------------------------------------|--------------------------------|
| 1 | 13,624.04 |
| 2 | 11,149.46 |
| 3 | 9,106.06 |
| 4 | 7,496.90 |
### Step 1: Determine how y values change
To check if the data might be linear or exponential, we can examine the differences between the values over the years:
1. Calculate the first differences (i.e., f(x) values minus the previous f(x) value):
- Year 2 - Year 1: \( 11,149.46 - 13,624.04 = -2,474.58 \)
- Year 3 - Year 2: \( 9,106.06 - 11,149.46 = -2,043.40 \)
- Year 4 - Year 3: \( 7,496.90 - 9,106.06 = -1,609.16 \)
2. These differences are not constant, which suggests it is not a linear function.
3. Next, we can calculate the ratios of the values:
- Year 2 / Year 1: \( 11,149.46 / 13,624.04 \approx 0.818 \)
- Year 3 / Year 2: \( 9,106.06 / 11,149.46 \approx 0.817 \)
- Year 4 / Year 3: \( 7,496.90 / 9,106.06 \approx 0.823 \)
4. These ratios are approximately constant, which suggests that an exponential function might fit the data well.
### Conclusion
Based on the analysis:
**Exponential function** would best fit the data because as **x** increases, the **y** values change **multiplicatively**. The **common ratio/multiplier/base** of this function is approximately **0.82**.
### Final Answer
Exponential function would best fit the data because as x increases, the y values change multiplicatively. The common ratio/multiplier/base of this function is approximately 0.82.
| Number of Years Since Investment Made, x | Value of Investment ($), f(x) |
|-------------------------------------------|--------------------------------|
| 1 | 13,624.04 |
| 2 | 11,149.46 |
| 3 | 9,106.06 |
| 4 | 7,496.90 |
### Step 1: Determine how y values change
To check if the data might be linear or exponential, we can examine the differences between the values over the years:
1. Calculate the first differences (i.e., f(x) values minus the previous f(x) value):
- Year 2 - Year 1: \( 11,149.46 - 13,624.04 = -2,474.58 \)
- Year 3 - Year 2: \( 9,106.06 - 11,149.46 = -2,043.40 \)
- Year 4 - Year 3: \( 7,496.90 - 9,106.06 = -1,609.16 \)
2. These differences are not constant, which suggests it is not a linear function.
3. Next, we can calculate the ratios of the values:
- Year 2 / Year 1: \( 11,149.46 / 13,624.04 \approx 0.818 \)
- Year 3 / Year 2: \( 9,106.06 / 11,149.46 \approx 0.817 \)
- Year 4 / Year 3: \( 7,496.90 / 9,106.06 \approx 0.823 \)
4. These ratios are approximately constant, which suggests that an exponential function might fit the data well.
### Conclusion
Based on the analysis:
**Exponential function** would best fit the data because as **x** increases, the **y** values change **multiplicatively**. The **common ratio/multiplier/base** of this function is approximately **0.82**.
### Final Answer
Exponential function would best fit the data because as x increases, the y values change multiplicatively. The common ratio/multiplier/base of this function is approximately 0.82.