rissaunnie
  • rissaunnie
urgent FAN AND MEDAL a country's population in 1991 was 147 million. In 1998 it was 153 million what's the population in 2017
Algebra
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
are you supposed to use exponential growth as in \[P(t)=P_0e^{rt}\]?
anonymous
  • anonymous
or can you do something easier?
rissaunnie
  • rissaunnie
its like P=Ae^kt

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anonymous
  • anonymous
we can do it that way if you like take a bit of time
rissaunnie
  • rissaunnie
sure
anonymous
  • anonymous
we can also do it a quick way using just the numbers given lets to it that way first
rissaunnie
  • rissaunnie
alright
anonymous
  • anonymous
in 7 year (from 1991 to 1998) it went from 147 to 153 you can model it as \[P(t)=147\times \left(\frac{153}{147}\right)^{\frac{t}{7}}\]
anonymous
  • anonymous
then if you want to know what it is 26 years later, from 1991 to 2017 replace \(t\) by \(26\) and get this http://www.wolframalpha.com/input/?i=147%28153%2F147%29^%2826%2F7%29
anonymous
  • anonymous
would you like to do it the slow way now?
anonymous
  • anonymous
set it up as \[P=Ae^{kt}\] and find \(k\) ?
anonymous
  • anonymous
you might have to do it on a test or sommat, i don't know it is up to you
rissaunnie
  • rissaunnie
since its timed i might use wolfram i just dont know how to enter the problem into it
rissaunnie
  • rissaunnie
thats y i didnt use it before
anonymous
  • anonymous
you can use the link i sent
anonymous
  • anonymous
otherwise we have to set \[153=147e^{9k}\] and solve for \(k\) then make the equation and evaluate at \(t=26\)
HeyItsZia1
  • HeyItsZia1
Substitute the original population and the model becomes \[P = 147e^{kt}\] 7 years later the population becomes 153. Substitute P with 153. \[153 = 147e^{7k}\] Now solve for k. \[153 = 147e^{7k}\] and t = 16
anonymous
  • anonymous
that is the second way do you know how to solve \[153=147e^{7k}\] for \(k\)?
rissaunnie
  • rissaunnie
no
anonymous
  • anonymous
do you want to know or are you satisfied with the answer above?

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