Students per Computer In the early years of microcomputers, school districts could not afford to buy a computer for every student. As the price of computers decreased, more and more school districts have been able to attain this goal. The following table lists numbers of students per computer during these early years.

year 1983 1985 1987 1989

students/computer 125 50 32 22

year 1991 1993 1995 1997

students/computer 18 14 10 6

(a) Make a scatterplot of the data. Would a straight line model the data accurately? Explain.

No, a straight line model would not model the data correctly, as the initially the fall is more rapid that later.

(b) Discuss how well the formula models these data, where S represents the students per

computer and y represents the year.

S = 125 ,y>or equal to 1983

1+.7(y-1983)

Using above formula, for

Year

Actual Students/computer

Modeled Students/computer

1983

125

125

1985

50

52

1987

32

33

1989

22

24

1991

18

19

1993

14

16

1995

10

13

1997

6

12

Thus the model prediction is quite close to the actual values.

 (c) In what year does the formula reveal that there were about 17 students per computer?

Solving,

17 = 125/(1+0.7(y-1983))

Or, 1+0.7(y-1983) = 125/17 = 7.3529

Or, y = 1983+(7.3529-1)/0.7 = 1992

So, in 1992, there were 17 students per computer.

AIDS Cases From 1993 to 2003 the cumulative number N of AIDS cases in thousands can be approximated by N=-2xsquare+76x+430, where x=0 corresponds to the year 1993.

Year 1993 1995 1997 1999 2001 2003

cases 422 565 677 762 844 930

(a) Use the equation to find N for each year in the table.

For 1993, putting x=0, N = -2*0^2+76*0+430 = 430

For 1995, putting x=2, N = -2*2^2+76*2+430 = 574

For 1997, putting x=4, N = -2*4^2+76*4+430 = 702

For 1999, putting x=6, N = -2*6^2+76*6+430 = 814

For 2001, putting x=8, N = -2*8^2+76*8+430 = 910

For 2003, putting x=10, N=-2*10^2+76*10+430 = 990

(b) Discuss how well this equation approximates the data.

The equation approximates to a good degree initially, but for further dates from 1993, the approximation gradually becomes a little more off the actual number of cases.

(c) Rewrite the equation with the right side completely factored.

-2x^2 + 76x + 430

= -2x^2 + 86x - 10x + 430

= -2x(x-43) - 10(x-43)

= (-2x-10) (x-43)

= -2(x+5)(x-43)

(d) Use your equation from part (c) to find N for each year in the table. Do your answers agree with those found in part (a)?

For 1993, putting x=0, N=-2(0-43)(0+5)= 430

For 1995, putting x=2, N=-2(2-43)(2+5) = 574

For 1997, putting x=4, N=-2(4-43)(4+5) = 702

For 1999, putting x=6, N=-2(6-43)(6+5) = 814

For 2001, putting x=8, N=-2(8-43)(8+5) = 910

For 2003, putting x=10, N=-2(10-43)(10+5) = 990

Thus, answers agree with those found in part (a).