Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Review Exercises - Page 1125: 52

Answer

a. See table and explanations. b. $a_n=15.98(1.02)^{n-1}$ c. $28.95$ (million)

Work Step by Step

a. See table. The first column is the year, the second column is the population (in millions), the third column is the ratio of the population divided by that of the preceding year, and the last column is the ratio rounded to two decimal places. We can see that $r=1.02$, indicating that the population increase approximately geometrically. b. Let $n$ be the year after 1999. We can write the general term of the geometric sequence as $a_n=a_1r^{n-1}=15.98(1.02)^{n-1}$ c. For year 2030, we have $n=2030-1999=31$. Thus $a_{31}=15.98(1.02)^{31-1}\approx28.95$ (million)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.