In 150 spins, about how often do you expect to less than 6?

A. The decrease in Car 1 value is constant at 6000 from year to year. Its value can be described by a linear function.

The decrease in Car 2 value is a constant percentage of its current value from year to year. Its value can be described by an exponential function.

B. The slope of the Car 1 function is -6000, and the value is 32000 in year 1, so the value function for Car 1 can be written as

... f(x) = 32000 -6000(x-1)

... f(x) = 38000 -6000x

The value multiplier for Car 2 from year to year is 27455/32300 = 0.85, so the value function for Car 2 can be written as

... f(x) = 32,300·0.85^(x-1)

... f(x) = 38000·0.85^x

C. There is a significant difference in value after 5 years. Car 2 is worth more than double the value of Car 1 in year 5. That year, their values are ...

... car 1: $8000

... car 2: $16861