Question 1102026: The population of Adams Ville grew from 10,000 to 18,000 in five years. Assuming uninhibited exponential growth, what is the expected population in an additional two years
Found 3 solutions by josgarithmetic, stanbon, greenestamps:
Answer by josgarithmetic(39630)   (Show Source):
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The population of Adamsville grew from 10,000 to 18,000 in five years. Assuming uninhibited exponential growth, what is the expected population in an additional two years.
Model:: pop = a*b^t
------
At time 0 the population is 10000
i,e, 10,000 = a*b^0
So, a = 10,000
-----
At time = 5 years the pop is 18000
pop = 10000*b^5
18000 = 10000*b^5
1.8 = b^5
b = 8^(1/5) = 1.1247
----
Equation::
pop = 10,000*1.1247^t
----
In an additional 2 yrs, t = 7
Then pop = 10000*1.1247 = 22765 when rounded up
-------------
Cheers,
Stan H.
----------
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website! The solutions by both earlier tutors did some rounding off during their calculations, so they came up with good approximate answers.
To get a more exact answer, don't round off until the final answer, where of course you want a whole number.
The population grew from 10,000 to 18,000 in 5 years; the annual exponential growth rate was 
The population after 7 years at the same rate will be
To the nearest whole number, the population after the additional 2 years will be 22771.
|
|
|
