Question 1129963: Use the average growth rate between 1850 and 1950, which was about 0.7%, to find the approximate doubling time and to predict the population in 2050 of a growing suburban town (based on a 2000 population of 100,000). Round to the nearest year and person.
A.102 years; 2050 population=151,572
B.100 years; 2050 population=141,421
C.95 years; 2050 population=123,114
D.110 years; 2050 population=131,951
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the average growth rate between 1850 and 1950 is equal to .7% which is equal to .007.
the formula to use is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods
if the 2000 population is 100,000, then doubling it will result in 200,000.
the formula becomes 200,000 = 100,000 * (1 + .007) ^ n
divide both sides of the equation by 100,000 and simplify to get:
2 = 1.007 ^ n
take the log of both sides of the equation to get:
log(2) = log(1.007 ^ n) which is the same as log(2) = n * log(1.007).
solve for n to get:
n = log(2) /log(1.007) = 99.36719646, which is closer to 100 than any of the other selections.
note that rule of 70 would have gotten you 70 / .7 = 100.
at .7% annual interest rate for 50 years, the formula becomes:
f = 100,000 * (1.007) ^ 50.
this results in f = 141,734.
this is closer to 141,421 than any of the other selections.
your solution points to selection B.
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