From 2009 to 2014 the percent of students placing into Math 101 has increased from 13% to 25%. Meanwhile, the percent of students placing into Math 222 has dropped from 25% to 7% If both placements change at a linear rate, what will the percentage be when the percent of people placing into Math 101 is equal to the percent placing into Math 222?
say 100 total students to make it easy
let y = year - 2009 to make it easy
M101
2009 13
2014 25
s = m y + b
m = (25-13)/5 = 12/5
13 = (12/5) 0 + b so b = 13
s = (12/5)y + 13
M222
2009 25
2014 7
s = (-18/5)y + 25
so when is
(12/5)y + 13 = -18 y/5 + 25 ???
30 y/5 = 12
y = 2
so in 2009 + 2 = 2011
then s = (- 36 / 5) + 25
= (125-36)/5 = 17.8 percent
To find the percentage at which the number of students placing into Math 101 is equal to the percentage placing into Math 222, we need to calculate the rate of change for each placement and determine when they intersect.
Let's begin by calculating the rate of change for each placement:
Rate of change for Math 101: (25% - 13%) / (2014 - 2009) = 12% / 5 years
Rate of change for Math 222: (7% - 25%) / (2014 - 2009) = -18% / 5 years
Now, let's set up an equation to find the intersection point:
13% + (12%/5) * x = 25% + (-18%/5) * x
Here, x represents the number of years since 2009 when the percentages will be equal.
To solve this equation, let's isolate x:
(12%/5) * x + (18%/5) * x = 25% - 13%
(30%/5) * x = 12%
(6%/5) * x = 12%
x = (12% * 5) / 6% ≈ 10
Therefore, it will take approximately 10 years for the percentage of students placing into Math 101 to be equal to the percentage placing into Math 222. To find the percentage at that point, substitute x = 10 into either equation:
13% + (12%/5) * 10 = 15.6%
Therefore, the percentage of students placing into Math 101 when it is equal to the percentage placing into Math 222 will be approximately 15.6%.