Baxter is thinking about buying a car. The table below shows the projected value of two different cars for three years. Number of years 1 2 3 Car 1 (value in dollars) 18,500 17,390 16,346.60 Car 2 (value in dollars) 18,500 17,500 16,500 Part A: What type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? Explain your answer. (2 points) Part B: Write one function for each car to describe the value of the car f(x), in dollars, after x years. (4 points) Part C: Baxter wants to purchase a car that would have the greatest value in 10 years. Will there be any significant difference in the value of either car after 10 years? Explain your answer, and show the value of each car after 10 years. (4 points)

Part A: Linear function

Part B: Y = -1000×X + 19463

Y = -1076.7×X + 19602

Part C: There will be significant difference, car 1 will be worth 628  more than car 2 after 10 years

Step-by-step explanation:

The table for the value of the two cars with time are given as follows;

Years                               Car 1                                     Car 2                

1                                     18,500                                    18,500

2                                    17,390                                     17,500

3                                    16,500                                     16,346.6

Plotting the give data, we have;

From the plot of the data, the function that can be used to describe the values of both cars after a fixed number of years is a linear function

Part B

From the relation for linear regression formula, Y = b·X + a, we have ;

For car 1

ΣXY = 102780

(ΣX)² = 36

ΣX = 6

ΣY = 52390

ΣX² = 14

N  = 3

b = -1000

a = 19463

Y = -1000×X + 19463

For Car 2

ΣXY = 102539.8

(ΣX)² = 36

ΣX = 6

ΣY = 52346.6

ΣX² = 14

N = 3 .

b = -1076.7

a = 19602

Y = -1076.7×X + 19602

Part C:

For Car 1 after 10 years, we have;

Y = -1000×X + 19463

Y = -1000×10 + 19463 = 9,463

For Car 2 after 10 years, we have;

Y = -1076.7×X + 19602

Y = -1076.7×10 + 19602 = 8,835

Therefore, car 1 will be worth 9,463 and car 2 will be worth 8,835 therefore, car 1 will be worth (9,463 - 8,835) = 628  more than car 2 after 10 years.

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