Question

In 1993, the average family income was about $35,000, and in 2008 it was about $48,998. Let x represent 1993, x+1 represent 1994, and so on. Find values for a and b (rounded to one decimal place if necessary) so that y=aeᵇˣ models the data.

a) a=35,000, b=0.03
b) a=35,000, b=0.02
c) a=48,998, b=0.03
d) a=48,998, b=0.02

Answer 1

To fit the family income data to the function y=aeˣˆˆx, a would be $35,000 since it's the income for 1993 (x=0), and b would be calculated based on the income for 2008. After calculating and comparing options, a=35,000 and b=0.02 provide the best fit to the data.

Step-by-step explanation:

To model the average family income data with the function y = aeˣˆˆx, we need to find the values for a and b that fit the data for the years 1993 and 2008. Given that in 1993 (x=0), the income was about $35,000, and in 2008 (x=15), it was about $48,998, we can use these two points to solve for a and b.

For the year 1993, plugging in x=0, the equation becomes y = a. Since we know that y was $35,000, it follows that a = $35,000. This eliminates options (c) and (d) because they both have a as $48,998.

Next, for the year 2008, plugging in x=15, the equation becomes y = 35,000eˣ(15b). We know that y should be $48,998, so we need to solve for b using this information. Doing the calculations (which involve logarithms and algebra).

we find which value of b gives the closest approximation to $48,998 when x=15 and a is confirmed as $35,000. In this case, option (b) with b = 0.02 provides the correct approximation, rounding to one decimal place if necessary.