Sec 1.4:

Sec 1.4:
Answer to problem #18
In this problem the book is using p as the independent variable ( there for the p-values are
on the x – axis and q values are on the y – axis.)
Since we know the q – intercept is 290 and the slope is 40(the average rate of change)
then q = 40t + 290. To evaluate the amount of pasta in 2005 you need to let t = 15 and
find q(15) = 40(15) + 290 = 890 million pounds
Answer to problem #48
In aligning data you get 1989 will be t = 0, 1994 will be t = 5, 1999 will be t = 10.
In modeling the information from 1989 to 1994 you will use the following points to
create a linear equation. ( 0, 100000) and (5, 85000) find the slope and then the linear
equation E – E1 = m(t – t1) to get E = -3000t + 100,000.
To model information from 1994 to 1999 you will use the points ( 5, 85000) and
(10, 88000) , find the slope and the equation E = 600t + 82000.
Now create the piecewise function E(t) to be the following:
E (t ) =
−3000 t +100000
600 t +82000
0 ≤t ≤5
5≺ t ≤10
The command for Excel would be as follows:
=If(A2<=5,-3000*a2+100000,600*a2+82000)
To evaluate 1995 then t = 6 and E(6) = 600(6) + 82000= 85600 students