anonymous
  • anonymous
The present population, P, in thousands, of a town is related to the initial population of the town at time t = 0 by the formulaP = 185(2)(0.05t)where t is measured in years. The town was populated in 1880. What was the population (to the nearest 1,000) in the year 2000?
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is formula 185*2*0.05t
anonymous
  • anonymous
?
anonymous
  • anonymous
no im sorry the formula is 185(2)^(0.05t)

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anonymous
  • anonymous
I think you mean: P(t) = \[185(2)^{(0.05t)}\] so if you subtract 2000-1880 to get t then substitute that into the equation.
anonymous
  • anonymous
t should be 120 120*.05 = 6 2^6 = 64 185*64 = 11840 people
anonymous
  • anonymous
or 12,000 people if you round
anonymous
  • anonymous
okay just to double check your answer, heres additional informatio. the population of the year 1880 was 185,000
anonymous
  • anonymous
then in that case set 1880= 185(2)^(0.05t) and solve for t. Add 120 to the t value and substitute that value back into the P(t)
anonymous
  • anonymous
and what did u come up with because im stuck
dumbcow
  • dumbcow
from 1880 to 2000, 120 years have passed so this will be the t_value you use to evaluate the population in year 2000 \[P = 185*2^{.05*120}\] \[P = 185*2^{6}\] \[P = 11,840\] since this is in thousands add 3 zeroes
anonymous
  • anonymous
wouldnt you add 11,840 to the population 185,000 because thats how much the popultion increased over 120 years?
dumbcow
  • dumbcow
no because we left the 185 in the equation their initial population is already factored in the increase would be 11,840 - 185
dumbcow
  • dumbcow
in other words the equation gives population at any time t not just the increase
anonymous
  • anonymous
damn. well then maybe idid it wrong and maybe the population of 1880 isnt 185,000 .....ughh im confusing myself.
anonymous
  • anonymous
??????????????????
dumbcow
  • dumbcow
If what you posted above is correct then yes when t=0 in 1880 the population is 185,000

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