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In 2006, there were 750 students enrolled in an online training program, and in 2010, there were 974 students enrolled in the pro

In 2006, there were 750 students enrolled in an online training program, and in 2010, there
were 974 students enrolled in the program. Let y be enrollment in the year x, where x = 0
represents the year 2006.
(a) Which of the following linear equations could be used to predict the enrollment y in a given
year x, where x = 0 represents the year 2006? Explain/show work.

A. y = 750x + 56
B. y = 56x + 224
C. y = 224x + 750
D. y = 56x + 750


(b) Use the equation from part (a) to predict the enrollment for the year 2013. Show work.


(c) Fill in the blanks to interpret the slope of the equation: The rate of change of enrollment with
respect to time is ______________________ per ________________. (Include units of
measurement.)
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3 Answers

two points in standard (x,y) configuration (0,750) and (4,974)
 
(a) find the slope :
     m=rise/run=(974-750)/(4-0)=224/4=56
 
     find the y-intercept (what is y when x=0), the first point
 
     Hence the equation is : y=56x+750
 
(b) what is the value of y when x = 2013-2006=7
     y = 56(7)+750=1142
 
( c)  students, year
 
(a) From 2006 to 2010, that is in 4 years,  the number of students increased from 750 to 974. 
increase in number of students per year  = (974 - 750) / 4 = 56
 
 If y be enrollment in the year x, where x = 0 represents the year 2006, we get the equation as
 
y = 56x + 750
 
To verify it, put x = 0 which gives y = 750 that is the enrollment in year 2006
x=4 which gives y = 974 that is the enrollment in the year 2010
 
(b) If x=0 for year 2006, x will be 7 for 2013. To find the enrollment in year 2013, we replace x by 7 in the above equation
 
y = 56*7 + 750 = 392+750 =1142
 
(c) As explained above, the rate of change of enrollment with respect to time is 56 students per year
A linear equation has the general form y = mx + b, where m is the slope and b is the y-intercept.  To determine the equation of a line, you need to know either the slope and one point and two points on the line.  The problem gives you two point: (2006,750) and (2010,974).  The problem tells you to use x=0 for 2006.  Hence the two points are (0,750) and (4,974).  The slope is:
 
m = (974-750)/(4-0) = 224/4 = 56
 
The y-intercept, b,occurs when x=0, so b= 750.  Hence the equation for the line is:
 
y = 56x + 750
 
2013-2006 = 7, so just plus x=7 into the equation above to predict enrollment is 2013.
 
The slope, m, is 56 which equates to an increase of 56 students enrolled each year
 
 

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