7% = .07 , not .7 like you had
That was your major mistake.
a) equation:
n(t) = 1900 e^(.07t)
b) if 2005 ---> t = 0
then 2010 ---> t = 5
n(5) = 19000 e^(.07*5) = 26962
I don't understand how you could possible think that 5 foxes would be correct.
there were 19000 foxes and they increased.
Didn't 5 sound rather "unreasonable" to you ?
c) , so we want n(t) = 25000
19000 e^(.07t) = 25000
e^ .07t = 1.315789
take ln of both sides and use log rules
.07t ln e = ln 1.315789
.07t = .274437
t = 3.920.. or appr 3.9 years
I know I can use the population growth formula. here is the question; The fox population in a certain region has a relative growth rate of 7% per year. It is estimated that the population in 2005 was 19,000.
(a) Find a function
n(t) = n0e^rt
that models the population t years after 2005.
n(t) = 1900e^(0.7t)
(b) Use the function from part (a) to estimate the fox population in the year 2010. (Round your answer to the nearest whole number.)
5
foxes
(c) After how many years will the fox population reach 25,000? (Round your answer to one decimal place.)
29 yr
so this is what I got but all 3 of my results were incorrect. can somebody show me step by step? I really am trying to know how to do this example.
1 answer