Answer
to Question Two: Is there a linear model that "fits"
the data?
(a.)
Using the TI-83 calculator, we enter the gas prices in L1 (these
will be our x-variables) and enter the net profits in L2 (these
will be our y-variables). We then perform a Linear Regression
between the two variable (L1,L2). [To perform the linear regression
go to STAT, CALC, LinReg (ax+b), enter, enter] This will give
you the following equation: y = -64.2x+1039.4
This
is a picture of the "line of best fit."
(b.) How well
does this function model our data? Let's test the points we were
given and compare the actual y-value to the predicted y-value.
| actual net income |
949.56 |
956.02 |
962.48 |
975.94 |
| predicted net income |
949.52 |
955.94 |
962.36 |
968.78 |
| gas price |
1.4 |
1.3 |
1.2 |
1.1 |
You can
see that the model does an excellent job predicting the actual
values of Tom's net income as it depends on the gas prices.
What
can we conclude from this model? We can see that as the price
of gas decreases, it is as if Tom is getting a raise. Therefore,
it is worth his while to shop around for the best gas prices he
can find (furthermore, it would be worth his time to lobby again
OPEC and for the United States to drill for their own oil in Alaska)
(c.) What would
the gas price have to be in order for Tom to meet his monthly
needs?
Let y = 1000
and solve for x.
1000=-64.2 x+
1039.4
x= .613
So, gas would
have to sell for $.60 in order for Tom to meet his monthly income
goal.