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
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?
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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 ".
