A rule of thumb used by car dealers is that the trade-in value of a car decreases by 30% each year. That is, the value at the end of any year is 70% of its value at the beginning of the year
A.Suppose that you own a car whose trade-in value is presently $2350. How much will it be worth one year from now? Two years from now? Three years from now?
We learned to use the equation y=KM^x+b
I just don’t know what to plug in
According to the rule stated in your first paragraph,
value at end of 1 year = 2350(.7) = ...
value at end of 2 yrs = 2350(.7)^2 = ...
value at end of 3 yrs. = 2350(.7)^3 = ...
To use the equation y = KM^x+b, we need to determine the values of K, M, and b based on the given information.
In this case, the information given is that the trade-in value of a car decreases by 30% each year. So, the multiplier M is 0.7 (since each year the car's value decreases to 70% of its previous value).
Now, let's determine the initial value (y) and the year (x) for each scenario given.
Scenario 1: One year from now (x = 1)
Given: Trade-in value = $2350
We want to find the value one year from now, so we can plug in the values:
2350 = K * 0.7^1 + b
Scenario 2: Two years from now (x = 2)
Given: Trade-in value = $2350
We want to find the value two years from now, so we can plug in the values:
2350 = K * 0.7^2 + b
Scenario 3: Three years from now (x = 3)
Given: Trade-in value = $2350
We want to find the value three years from now, so we can plug in the values:
2350 = K * 0.7^3 + b
To solve these equations and find the values of K and b, you can use algebraic methods such as substitution or elimination. Once you find K and b, you can then calculate the trade-in value at each point in time (one year from now, two years from now, and three years from now) by plugging in the respective values of x.
Please note that without knowing the specific values of K and b, I am unable to provide the exact trade-in values for each scenario.