The table lists data regarding the average salaries of several professional athletes in the years 1991 and 2001.
year average salary
1991 $263,000
2001 $1,430,000
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
Could someone please check my answers
a)S(x)=116,700x+146,300
b) in 2005 $1,780,100
in 2010 $2,363,600
no, if you are going to use the actual year for the value of x, I got
S(x) = 116700x - 232086700
then for 2005
S(2005) = 116700(2005) - 232086700
= 1896800
and S(2010) = 116700(2010) - 232086700
= 2480300
btw, the two given ordered pairs satisfy my equation
To find a linear function that fits the data, we can use the formula for a linear equation: y = mx + b, where "x" represents the year and "y" represents the average salary.
First, we need to find the slope (m) and the y-intercept (b) using the given data points.
Using the two data points (1991, $263,000) and (2001, $1,430,000), we can calculate the slope (m):
m = (change in y) / (change in x) = (1,430,000 - 263,000) / (2001 - 1991)
m = 1,167,000 / 10
m = 116,700
Now we can substitute the slope (m) into the equation y = mx + b and solve for the y-intercept (b).
Using the point (1991, $263,000) as (x, y) = (1991, 263,000):
263,000 = 116,700 * 1991 + b
263,000 = 232,238,700 + b
b = 263,000 - 232,238,700
b = -231,975,700
Therefore, the linear function that fits the data is:
S(x) = 116,700x - 231,975,700
Now let's use this function to predict the average salaries in 2005 and 2010:
For the year 2005:
x = 2005
S(2005) = 116,700 * 2005 - 231,975,700
S(2005) = 233,185,500 - 231,975,700
S(2005) = $1,209,800
For the year 2010:
x = 2010
S(2010) = 116,700 * 2010 - 231,975,700
S(2010) = 234,987,000 - 231,975,700
S(2010) = $3,011,300
Therefore, the predicted average salary in 2005 is $1,209,800, and in 2010 is $3,011,300.