the table lists data regarding the average salaries of several profssional athletes in the years 1991 and2001
a) use the data points to find a linear function that fits the data
b) use the function to predict the average salary in 2005 and 2010 in 1991 the salry was $1440,000 and in 2001 the salary was $257000 Sx= and the salary in 2005 is and the salary in 2010 will be
To find a linear function that fits the data, we can use the formula for the equation of a straight line, which is:
y = mx + b,
where "y" represents the dependent variable (average salary in this case), "x" represents the independent variable (year), "m" is the slope of the line, and "b" is the y-intercept.
Based on the given data points:
- In 1991, the salary was $1,440,000
- In 2001, the salary was $257,000
We can assign these values to corresponding variables:
- For 1991: x₁ = 1991 and y₁ = 1,440,000
- For 2001: x₂ = 2001 and y₂ = 257,000
To calculate the slope (m), we use the formula:
m = (y₂ - y₁) / (x₂ - x₁).
Plugging in the values, we get:
m = (257,000 - 1,440,000) / (2001 - 1991)
= -1,183,000 / 10
= -118,300.
Now, let's find the y-intercept (b). We can plug the values of one of the given data points into the equation and solve for b. Let's use the point (x₁, y₁):
1,440,000 = -118,300 * 1991 + b
Simplifying the equation:
b = 1,440,000 + 118,300 * 1991
= 1,440,000 + 235,057,300
= 236,497,300.
So, the linear function that fits the data is:
y = -118,300x + 236,497,300.
Now, let's move on to predicting the average salary in 2005 and 2010 using this function.
In 2005, let x = 2005:
y = -118,300 * 2005 + 236,497,300.
Calculating this expression will give us the predicted average salary in 2005.
Similarly, for 2010, let x = 2010:
y = -118,300 * 2010 + 236,497,300.
Calculating this expression will give us the predicted average salary in 2010.