Cost of Tuition In 2000, the average cost of tuition
and fees at private four-year colleges was $16,200,
and in 2005 it was $20,100. Sketch a line that passes
through the points (2000, 16200) and (2005, 20100).
To sketch a line that passes through the points (2000, 16200) and (2005, 20100), you can follow these steps:
Step 1: Determine the slope (m) of the line.
The slope (m) is calculated using the formula:
m = (y2 - y1) / (x2 - x1)
Given that the points are (2000, 16200) and (2005, 20100), we can substitute the values into the formula to find the slope:
m = (20100 - 16200) / (2005 - 2000)
m = 3900 / 5
m = 780
Step 2: Determine the y-intercept (b) of the line.
The y-intercept (b) is the value of y when x is equal to 0. To find it, we can use one of the given points.
Using the point (2000, 16200), we can substitute the x and y values into the equation y = mx + b, and solve for b:
16200 = 780 * 2000 + b
16200 = 1560000 + b
b = 16200 - 1560000
b = -1543800
Step 3: Write the equation of the line.
Using the slope-intercept form of a linear equation (y = mx + b), we can substitute the slope (m) and the y-intercept (b) to write the equation of the line:
y = 780x - 1543800
Now, you can use this equation to sketch the line passing through the points (2000, 16200) and (2005, 20100) on a graph.
To sketch a line passing through the points (2000, 16200) and (2005, 20100), we can use the slope-intercept form of a linear equation: y = mx + b.
First, let's find the slope (m):
m = (y2 - y1) / (x2 - x1)
= (20100 - 16200) / (2005 - 2000)
= 1900 / 5
= 380
Now, let's find the equation of the line by substituting the slope and one of the points into the equation:
y = mx + b
16200 = 380 * 2000 + b
Solving for b:
b = 16200 - 380 * 2000
b = 16200 - 760000
b = -743800
The equation of the line passing through the points is:
y = 380x - 743800
Now we can sketch this line on a graph.