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April 17, 2015

Posted by **Golkhadi nep** on Friday, June 4, 2010 at 5:36pm.

(a) the number of pairs of cuckoos in the area in the summer of 2009;

(b) the number of pairs of cuckoos in the area in the summer of 2020

- Math -
**Ariel**, Friday, June 4, 2010 at 5:48pmso my guess is that you would start with what ya got

2008 560-4%

2009 559.96-4%

2010 559.92-4%

(continue til you get to 2020)

sooo;

a.) 559.92

b.) 559.52

- Math -
**MathMate**, Friday, June 4, 2010 at 7:04pmIf the population declines at 4%, it means that the surviving population is 96% of the previous year.

In the summer of 2008, there were 560 pairs. So in the summer of 2009, there were 560*(96/100)

= 537.6 pairs (theoretically)

= 538 pairs

In the summer of 2009, the population declined again to 96% of the theoretical population of 2008, thus

537.6*(96/100)

=516.1 pairs (theoretically)

= 516 pairs (rounded)

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