Answer :
To determine which of the given functions best models the data, we need to compare how well each function fits the provided data points. One common approach to doing this is by calculating the residuals (the differences between the observed values and the values predicted by the model) and then summing the squares of these residuals. The model with the smallest sum of squared residuals is deemed the best fit.
Here's the detailed process step-by-step:
### Step 1: Define the Data Points
We have the following data points from the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of days } (x) & \text{Estimated number of bees } (y) \\ \hline 0 & 10000 \\ 10 & 7500 \\ 20 & 5600 \\ 30 & 4200 \\ 40 & 3200 \\ 50 & 2400 \\ \hline \end{array} \][/tex]
### Step 2: Define the Proposed Models
We have four proposed models:
1. [tex]\( y = 9958 (0.972)^x \)[/tex]
2. [tex]\( y = (0.972)(9958)^x \)[/tex]
3. [tex]\( y = 9219x - 150 \)[/tex]
4. [tex]\( y = -150x + 9219 \)[/tex]
### Step 3: Calculate the Residuals For Each Model
- Model 1: [tex]\( y = 9958(0.972)^x \)[/tex]
1. For [tex]\( x = 0 \)[/tex]: [tex]\( y = 9958 (0.972)^0 = 9958 \rightarrow \text{residual} = 10000 - 9958 \)[/tex]
2. For [tex]\( x = 10 \)[/tex]: [tex]\( y = 9958 (0.972)^{10} \approx 7263.67 \rightarrow \text{residual} = 7500 - 7263.67 \)[/tex]
3. For [tex]\( x = 20 \)[/tex]: [tex]\( y = 9958 (0.972)^{20} \approx 5294.92 \rightarrow \text{residual} = 5600 - 5294.92 \)[/tex]
4. For [tex]\( x = 30 \)[/tex]: [tex]\( y = 9958 (0.972)^{30} \approx 3857.89 \rightarrow \text{residual} = 4200 - 3857.89 \)[/tex]
5. For [tex]\( x = 40 \)[/tex]: [tex]\( y = 9958 (0.972)^{40} \approx 2812.18 \rightarrow \text{residual} = 3200 - 2812.18 \)[/tex]
6. For [tex]\( x = 50 \)[/tex]: [tex]\( y = 9958 (0.972)^{50} \approx 2048.72 \rightarrow \text{residual} = 2400 - 2048.72 \)[/tex]
- Model 2: [tex]\( y = (0.972)(9958)^x \)[/tex]
Note: Given the exponential nature, we'd expect a very rapid increase or decrease, and comparing the actual residuals could be impractical because 9958^x grows very quickly.
Based on logical consideration, such a model is less likely to fit the given data since the population should decay, not grow rapidly.
- Model 3: [tex]\( y = 9219x - 150 \)[/tex]
This is a linear model suggesting a constant rate of increase per day. Given the data trend showing a decrease, it is unlikely to fit well.
- Model 4: [tex]\( y = -150x + 9219 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]: [tex]\( y = -150(0) + 9219 = 9219 \rightarrow \text{residual} = 10000 - 9219 \)[/tex]
2. For [tex]\( x = 10 \)[/tex]: [tex]\( y = -150(10) + 9219 = 7719 \rightarrow \text{residual} = 7500 - 7719 \)[/tex]
3. For [tex]\( x = 20 \)[/tex]: [tex]\( y = -150(20) + 9219 = 6219 \rightarrow \text{residual} = 5600 - 6219 \)[/tex]
4. For [tex]\( x = 30 \)[/tex]: [tex]\( y = -150(30) + 9219 = 4719 \rightarrow \text{residual} = 4200 - 4719 \)[/tex]
5. For [tex]\( x = 40 \)[/tex]: [tex]\( y = -150(40) + 9219 = 3219 \rightarrow \text{residual} = 3200 - 3219 \)[/tex]
6. For [tex]\( x = 50 \)[/tex]: [tex]\( y = -150(50) + 9219 = 1719 \rightarrow \text{residual} = 2400 - 1719 \)[/tex]
### Step 4: Sum of Squared Residuals
- Model 1 Residuals:
[tex]\[ \begin{align*} & (10000 - 9958)^2 + (7500 - 7263.67)^2 + (5600 - 5294.92)^2 + (4200 - 3857.89)^2 + \\ & (3200 - 2812.18)^2 + (2400 - 2048.72)^2 \end{align*} \][/tex]
- Model 4 Residuals:
[tex]\[ \begin{align*} & (10000 - 9219)^2 + (7500 - 7719)^2 + (5600 - 6219)^2 + (4200 - 4719)^2 + \\ & (3200 - 3219)^2 + (2400 - 1719)^2 \end{align*} \][/tex]
### Conclusion
Calculating and comparing the actual sum of squared residuals, Model 1 (exponential decay) should have smaller errors since it follows a decaying trend, capturing the overall pattern of the bee population reducing over time.
Therefore, the function that best models the data is:
[tex]\[ y = 9958 (0.972)^x \][/tex]
Here's the detailed process step-by-step:
### Step 1: Define the Data Points
We have the following data points from the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of days } (x) & \text{Estimated number of bees } (y) \\ \hline 0 & 10000 \\ 10 & 7500 \\ 20 & 5600 \\ 30 & 4200 \\ 40 & 3200 \\ 50 & 2400 \\ \hline \end{array} \][/tex]
### Step 2: Define the Proposed Models
We have four proposed models:
1. [tex]\( y = 9958 (0.972)^x \)[/tex]
2. [tex]\( y = (0.972)(9958)^x \)[/tex]
3. [tex]\( y = 9219x - 150 \)[/tex]
4. [tex]\( y = -150x + 9219 \)[/tex]
### Step 3: Calculate the Residuals For Each Model
- Model 1: [tex]\( y = 9958(0.972)^x \)[/tex]
1. For [tex]\( x = 0 \)[/tex]: [tex]\( y = 9958 (0.972)^0 = 9958 \rightarrow \text{residual} = 10000 - 9958 \)[/tex]
2. For [tex]\( x = 10 \)[/tex]: [tex]\( y = 9958 (0.972)^{10} \approx 7263.67 \rightarrow \text{residual} = 7500 - 7263.67 \)[/tex]
3. For [tex]\( x = 20 \)[/tex]: [tex]\( y = 9958 (0.972)^{20} \approx 5294.92 \rightarrow \text{residual} = 5600 - 5294.92 \)[/tex]
4. For [tex]\( x = 30 \)[/tex]: [tex]\( y = 9958 (0.972)^{30} \approx 3857.89 \rightarrow \text{residual} = 4200 - 3857.89 \)[/tex]
5. For [tex]\( x = 40 \)[/tex]: [tex]\( y = 9958 (0.972)^{40} \approx 2812.18 \rightarrow \text{residual} = 3200 - 2812.18 \)[/tex]
6. For [tex]\( x = 50 \)[/tex]: [tex]\( y = 9958 (0.972)^{50} \approx 2048.72 \rightarrow \text{residual} = 2400 - 2048.72 \)[/tex]
- Model 2: [tex]\( y = (0.972)(9958)^x \)[/tex]
Note: Given the exponential nature, we'd expect a very rapid increase or decrease, and comparing the actual residuals could be impractical because 9958^x grows very quickly.
Based on logical consideration, such a model is less likely to fit the given data since the population should decay, not grow rapidly.
- Model 3: [tex]\( y = 9219x - 150 \)[/tex]
This is a linear model suggesting a constant rate of increase per day. Given the data trend showing a decrease, it is unlikely to fit well.
- Model 4: [tex]\( y = -150x + 9219 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]: [tex]\( y = -150(0) + 9219 = 9219 \rightarrow \text{residual} = 10000 - 9219 \)[/tex]
2. For [tex]\( x = 10 \)[/tex]: [tex]\( y = -150(10) + 9219 = 7719 \rightarrow \text{residual} = 7500 - 7719 \)[/tex]
3. For [tex]\( x = 20 \)[/tex]: [tex]\( y = -150(20) + 9219 = 6219 \rightarrow \text{residual} = 5600 - 6219 \)[/tex]
4. For [tex]\( x = 30 \)[/tex]: [tex]\( y = -150(30) + 9219 = 4719 \rightarrow \text{residual} = 4200 - 4719 \)[/tex]
5. For [tex]\( x = 40 \)[/tex]: [tex]\( y = -150(40) + 9219 = 3219 \rightarrow \text{residual} = 3200 - 3219 \)[/tex]
6. For [tex]\( x = 50 \)[/tex]: [tex]\( y = -150(50) + 9219 = 1719 \rightarrow \text{residual} = 2400 - 1719 \)[/tex]
### Step 4: Sum of Squared Residuals
- Model 1 Residuals:
[tex]\[ \begin{align*} & (10000 - 9958)^2 + (7500 - 7263.67)^2 + (5600 - 5294.92)^2 + (4200 - 3857.89)^2 + \\ & (3200 - 2812.18)^2 + (2400 - 2048.72)^2 \end{align*} \][/tex]
- Model 4 Residuals:
[tex]\[ \begin{align*} & (10000 - 9219)^2 + (7500 - 7719)^2 + (5600 - 6219)^2 + (4200 - 4719)^2 + \\ & (3200 - 3219)^2 + (2400 - 1719)^2 \end{align*} \][/tex]
### Conclusion
Calculating and comparing the actual sum of squared residuals, Model 1 (exponential decay) should have smaller errors since it follows a decaying trend, capturing the overall pattern of the bee population reducing over time.
Therefore, the function that best models the data is:
[tex]\[ y = 9958 (0.972)^x \][/tex]
