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Answer:
a) slope = -765 jobs per year
b) j = -765y +1582855 . . . . jobs in year y
c) 38,320 jobs in 2019
Explanation:
a) We can define the variables y and j to represent the year number and the number of jobs, respectively. Then we are given 2 points:
(y, j) = (2012, 43675) and (2017, 39850)
The slope of the relation is ...
m = Δj/Δy = (39850 -43675)/(2017 -2012) = -3825/5 = -765
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b) Using the point-slope form of the equation of a line, we can model the relation between years and jobs as ...
y -k = m(x -h) . . . . . . . line with slope m through point (h, k)
For m = -765 and (h, k) = (2012, 43675), the model gives ...
j -43675 = -765(y -2012)
j = -765y +1582855 . . . . . . . rearrange to slope-intercept form
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c) For y = 2019, the number of jobs is modeled to be ...
j = -765(2019) +1582855 = -1544535 +1582855
j = 38,320 . . . . . jobs in 2019
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Additional comment
Quite often for problems of this sort, we let the time variable represent "years after <some reference year>". Here, it would be convenient to use 2012 as the reference year. The the equation would be j = -765y +43675, and the "j-intercept" would have the sensible value of 43675 at "year 0" (or 2012).
The equation we show is technically correct, but has no meaning for year values less than 2012. Especially, we know for sure there were not 1.6 million jobs in information in Indiana at the beginning of the Common Era. Indiana didn't exist until 1816.