Asked by george
The population of a Midwestern city decays exponentially. If the population decreased from 900,000 to 800,000 from 2003 to 2005, what will be the population in 2008?
I thought it was 7542 but it is wrong can someone please help me thank you
I thought it was 7542 but it is wrong can someone please help me thank you
Answers
Answered by
Reiny
let the rate of decay be r, where r is a decimal
so letting 2003 correspond with t = 0 , then
2005 corresponds with t = 2
900,000 r^2 = 800,000
r^2 = 8/9
r = √(8/9) = appr .9428
So count = 9000(.0428)^t
in 2008 , t = 5
count = 900,000(.9428)^5 = appr 670,442
or 670,000 rounded to the nearest thousand
so letting 2003 correspond with t = 0 , then
2005 corresponds with t = 2
900,000 r^2 = 800,000
r^2 = 8/9
r = √(8/9) = appr .9428
So count = 9000(.0428)^t
in 2008 , t = 5
count = 900,000(.9428)^5 = appr 670,442
or 670,000 rounded to the nearest thousand
Answered by
Scott
pop = 900000 * k^t
800000 = 900000 * k^2
8/9 = k^2 ... k = 2√2 / 3
p = 900000 * (2√2 / 3)^5
800000 = 900000 * k^2
8/9 = k^2 ... k = 2√2 / 3
p = 900000 * (2√2 / 3)^5
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