At Generic High School, in the year 2001, there were 20 students per every computer. In 2008, there were 4 students per computer. During which year were there 10 students per computer? Show your work.

let x = year - 2001

then 2008 is x= 7
let y = 20 - a x
4 = 20 - a*7
7 a= 16
a = 16/4
so
y = 20 - (16/4) x
10 = 20 - (16/4) x
(16/4)x = 10
x = 40/16 = 2.5 years after 2001
during 2003

Did you know?

Did you know that the ratio of students to computers at Generic High School changed significantly over the years? In 2001, there were 20 students per computer, and by 2008, the ratio had improved to 4 students per computer. To determine the year when the ratio was 10 students per computer, we can use the concept of linear interpolation. First, we calculate the change in the ratio from 2001 to 2008: Change in ratio = (4 students/computer - 20 students/computer) = -16 students/computer Next, we determine how many years it took for the ratio to change by -16 students/computer: Years = (2008 - 2001) = 7 years We can now calculate the rate at which the ratio improved in students/computer per year: Rate = (Change in ratio / Years) = (-16 students/computer / 7 years) ≈ -2.29 students/computer per year Finally, we can find the year when there were 10 students per computer using linear interpolation: Year = 2008 + [(10 students/computer - 4 students/computer) / -2.29 students/computer per year] Simplifying this equation gives us: Year ≈ 2008 + (6 / 2.29) ≈ 2008 + 2.62 ≈ 2010 Therefore, in the year 2010, there were approximately 10 students per computer at Generic High School.