No doubt your text advises how to arrive at a best-fit exponential model. The two graphing calculators I used gave different answers. The first one modeled the function as
y = 398.494(0.50526)^x
For x = 7, this model gives y = 3.35
The second calculator, a TI-84 work-alike, modeled the function as
y = 422.110(0.48896)^x
For x=7, this model gives y = 2.82
Using the first and last data points only, the model is
y = 200(0.06)^((x-1)/4)
For x=7, this model gives y = 2.94
Using yet another calculator to do linear regression on the log of y, the function is modeled by
y = 10^(2.62543-0.31073x)
which is another way to write the function produced by the TI-84 work-alike.
So, the technology I used gives a couple of reasons to choose
A. 2.86
and a couple of reasons to choose
C. 2.95
Of the models shown here, the ones with the lowest mean absolute deviation and the least squared error were the ones that produced the highest values of y for x=7, suggesting 2.95 is the best choice.
It would be advisable to use the procedure described in your text.
y = 398.494(0.50526)^x
For x = 7, this model gives y = 3.35
The second calculator, a TI-84 work-alike, modeled the function as
y = 422.110(0.48896)^x
For x=7, this model gives y = 2.82
Using the first and last data points only, the model is
y = 200(0.06)^((x-1)/4)
For x=7, this model gives y = 2.94
Using yet another calculator to do linear regression on the log of y, the function is modeled by
y = 10^(2.62543-0.31073x)
which is another way to write the function produced by the TI-84 work-alike.
So, the technology I used gives a couple of reasons to choose
A. 2.86
and a couple of reasons to choose
C. 2.95
Of the models shown here, the ones with the lowest mean absolute deviation and the least squared error were the ones that produced the highest values of y for x=7, suggesting 2.95 is the best choice.
It would be advisable to use the procedure described in your text.