Question 1123445
Let 2015 be t=0, 2016 be t=1, ...
It depreciated $11000 in 2 years, or $5500 a year
V(t)=36000-5500t function
2018 will be t=3
V(3)=36000-5500*3=$19500
It would appear not to be fair based on depreciation, especially since in 2018 can be considered year 3 in January or December. The problem isn't clear on when in the year the clock starts or stops
Since cars don't depreciate linearly but a great deal at first, if she were told this was a linear issue, she was not told the right thing.