SOLUTION: Suppose a population of pavement ants in Taylorsville, UT has an initial population of 2100 and 18 years later the population reaches 25000. Use an explicit exponential model to fi
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Question 1146528: Suppose a population of pavement ants in Taylorsville, UT has an initial population of 2100 and 18 years later the population reaches 25000. Use an explicit exponential model to find the rate of growth and common ratio for the pavement ant population.
Express the rate of growth as a percentage. Round to the nearest tenth.
r
=
%
Express the common ratio as a decimal. Round to the nearest thousandth.
1
+
r
= Found 2 solutions by Theo, rothauserc:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! the exponential growth formula is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is he interest rate per time period
n is the number of time periods.
in your problem, the formula becomes 25000 = 2100 * (1 + r) ^ 18
divide both sides of this equation by 2100 to get:
25000/2100 = (1 + r) ^ 18
take the 18th root of both sides of the equation to get:
(25000/2100) ^ (1/18) = 1 + r
subtract 1 from both sides of the equation to get:
(25000/2100) ^ (1/18) - 1 = r
solve for r to get:
r = .1475252801
multiply by 100 to get 14.75252801%.
that is the annual rate of growth.
replace r in the original equation to get:
25000 = 2100 * (1+.1475252801)^18
evaluate to get:
25000 = 25000
this confirms the annual rate of growth is correct.
the annual rate of growth is 14.75252801%
the common ratio sounds like it would be 1.1475252801.
each year, the population of pavement ants grows by that factor.
You can put this solution on YOUR website! The explicit exponential model equation is
:
P(n) = P(0) * (1 + r)^n, where P(0) is the initial population, r is growth rate, P(n) is the population after n years
:
Growth rate
:
25000 = 2100 * (1 + r)^18
:
(1 + r)^18 = 25000/2100 = 11.9048
:
1 + r = 11.9048^(1/18) = 1.1475
:
r = 0.1475
:
rate of growth = 14.8%
:
Common ratio
:
common ratio = 1.1475 is approximately 1.148
:
