SOLUTION: The population of a town grows at a rate of e^1.2t-2t people per year (where t is the number of years). At t=2 years, the town has 1500 people. Approximately by how many pe

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Question 1200292: The population of a town grows at a rate of e^1.2t-2t people per year (where t is the number of years). At t=2 years, the town has 1500 people. Approximately by how many people do the population grow between t=2 and t=5? What is the town’s population at t=5 year?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!

Nobody seems to want this, here my effort, for what its worth, I don't know
:
The population of a town grows at a rate of people per year (where t is the number of years).
At t=2 years, the town has 1500 people.

Approximately by how many people do the population grow between t=2 and t=5?
t=2, 1500, t=5, 1900. 1900 - 1500 = 400 increase from yr 2 to yr 5
What is the town’s population at t=5 year?
Approx 1900 people
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
The population of a town grows at a rate of e^1.2t-2t people per year (where t is the number of years).
At t=2 years, the town has 1500 people. Approximately by how many people do the population grow
between t=2 and t=5? What is the town’s population at t=5 year?
~~~~~~~~~~~~~~~~~~~~


My personal opinion is that this problem is worded and posed incorrectly.

More concretely, it is a soup of words, that are used incorrectly.


It is why I did not touch it and do not want to touch it. 


    If to read it literally, then it requires to solve a differential equation


         = ,


    where "p" is the population.



    But in this case,  the expression    is not  " the rate per year " : it is  simply  " the rate "

    or, if to be precisely accurate, " the instantaneous rate ".



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