Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.1 - Sequences - Exercise Set: 43

Answer

$2400$ cases $75$ cases.

Work Step by Step

$a_{n}= 75(2)^{n-1}$ No. of cases at the beginning of the sixth year $a_{6}= 75(2)^{6-1} $ $a_{6}= 75(2)^{5}$ $a_{6}= 75(32)$ $a_{6}= 2400$ No. of cases at the beginning of the year $a_{1}= 75(2)^{1-1} $ $a_{1}= 75(2)^{0}$ $a_{1}= 75(1)$ $a_{1}= 75$
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