Math 209 Learning Team Assignment Week 2 Aids Case

Directions:

Form a group of 2 to 4 people. Select someone to record the group’s responses for this activity. All members of the group should work cooperatively to answer the questions. If your instructor asks for your results, each member of the group should be prepared to respond.

 AIDS Cases

From 1993 to 2003 the cumulative number N of AIDS cases in thousands can be approximated by N = -2x2 + 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.

In 1993 (x = 0);   N(0)=-2(0)2+76(0)+430=430

In 1995 (x = 2);   N(2)=-2(2)2+76(2)+430=574

In 1997 (x = 4);   N(4)=-2(4)2+76(4)+430=702

In 1999 (x = 6);   N(6)=-2(6)2+76(6)+430=814

In 2001 (x = 8);   N(8)=-2(8)2+76(8)+430=910

In 2003 (x = 10);   N(10)=-2(10)2+76(10)+430=990

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

This equation shows us approximately on the average the increase in the many A.I.D.S. cases from 1993 to 2003. By using the formula provided in Part A, it shows how the many A.I.D.S. cases will continue to grow as the years pass by. The difference between the answers provided by the formula and the cases posted on the chart range from 1.86% to 6.06%. This range tells us that the formula gets us pretty close but is not perfect when compared to the actual number of cases posted.

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

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

(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)?

In 1993 (x = 0);   N(0) = -2(0-43) (0+5) = 430

In 1995 (x = 2);   N(2) = -2(2-43) (2+5) = 574

In 1997 (x = 4);   N(4) = -2(4-43) (4+5) = 702

In 1999 (x = 6);   N(6) = -2(6-43) (6+5) = 814

In 2001 (x = 8);   N(8) = -2(8-43) (8+5) = 910

In 2003 (x = 10);   N(10) = -2 (10-43) (10+5) = 990

These answers...