The table shows the amount of milk that Wisconsin dairy farms produced from 1937 to 1968. Which model best fits this data? Note that the year 1900 corresponds to x = 0.
Year
Milk Produced
(in billions of lbs.)
1937
12
1968
16
1995
20
(1 point)
Responses
y equals 0.138 x minus 6.819
Image with alt text: y equals 0.138 x minus 6.819
y equals negative 0.138 x minus 6.819
Image with alt text: y equals negative 0.138 x minus 6.819
y equals 0.138 x plus 6.819
Image with alt text: y equals 0.138 x plus 6.819
y equals negative 0.138 x plus 6.819
Image with alt text: y equals negative 0.138 x plus 6.819
The best model that fits this data is:
y equals 0.138 x minus 6.819
To determine which model best fits the data, we can use the given data points to calculate the slope and y-intercept of each equation and see which one aligns closest to the data.
We have three data points: (1937, 12), (1968, 16), and (1995, 20). We can use these points to calculate the slope and y-intercept for each equation and then compare them to the data.
Let's calculate using equation 1: y = 0.138x - 6.819
For the first data point (1937, 12):
y = 0.138 * 1937 - 6.819
y = 267.006 - 6.819
y ≈ 260.187
For the second data point (1968, 16):
y = 0.138 * 1968 - 6.819
y = 271.584 - 6.819
y ≈ 264.765
For the third data point (1995, 20):
y = 0.138 * 1995 - 6.819
y = 275.91 - 6.819
y ≈ 269.091
Now let's calculate using equation 2: y = -0.138x - 6.819.
For the first data point (1937, 12):
y = -0.138 * 1937 - 6.819
y = -267.006 - 6.819
y ≈ -273.825
For the second data point (1968, 16):
y = -0.138 * 1968 - 6.819
y = -271.584 - 6.819
y ≈ -278.403
For the third data point (1995, 20):
y = -0.138 * 1995 - 6.819
y = -275.91 - 6.819
y ≈ -282.729
Now let's calculate using equation 3: y = 0.138x + 6.819.
For the first data point (1937, 12):
y = 0.138 * 1937 + 6.819
y = 267.006 + 6.819
y ≈ 273.825
For the second data point (1968, 16):
y = 0.138 * 1968 + 6.819
y = 271.584 + 6.819
y ≈ 278.403
For the third data point (1995, 20):
y = 0.138 * 1995 + 6.819
y = 275.91 + 6.819
y ≈ 282.729
Lastly, let's calculate using equation 4: y = -0.138x + 6.819.
For the first data point (1937, 12):
y = -0.138 * 1937 + 6.819
y = -267.006 + 6.819
y ≈ -260.187
For the second data point (1968, 16):
y = -0.138 * 1968 + 6.819
y = -271.584 + 6.819
y ≈ -264.765
For the third data point (1995, 20):
y = -0.138 * 1995 + 6.819
y = -275.91 + 6.819
y ≈ -269.091
Now, let's compare the calculated y-values with the actual data points:
For equation 1: (1937, 12) ≈ (260.187, 12), (1968, 16) ≈ (264.765, 16), (1995, 20) ≈ (269.091, 20)
For equation 2: (1937, 12) ≈ (-273.825, 12), (1968, 16) ≈ (-278.403, 16), (1995, 20) ≈ (-282.729, 20)
For equation 3: (1937, 12) ≈ (273.825, 12), (1968, 16) ≈ (278.403, 16), (1995, 20) ≈ (282.729, 20)
For equation 4: (1937, 12) ≈ (-260.187, 12), (1968, 16) ≈ (-264.765, 16), (1995, 20) ≈ (-269.091, 20)
From the comparison, we can see that equation 3, y = 0.138x + 6.819, aligns closest with the given data points. Therefore, equation 3 is the best model that fits the data.
To determine which model best fits the data, we can use the given years and corresponding milk production values (in billions of lbs.) to find the equation that represents the relationship between the two variables.
We can start by identifying the slope of the line, which represents the rate of change in milk production over time.
Using the first and last data points, we can calculate the slope as follows:
slope = (change in milk production) / (change in years)
slope = (16 - 12) / (1968 - 1937)
slope = 4 / 31
The slope is approximately 0.129.
Next, we need to find the y-intercept, which represents the starting point of the line. We can use one of the data points and the slope to find the y-intercept.
Let's use the first data point (1937, 12):
12 = 0.129 * 1937 + b
12 = 249.573 + b
b ≈ -237.573
Therefore, the equation for the line that best fits the data is:
y = 0.129x - 237.573
Since none of the given models exactly match this equation, none of the provided options are the best fit for the data.