SOLUTION: A town’s population has been increasing at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070 people. Assume this trend continues

Question 1190102: A town’s population has been increasing at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070 people. Assume this trend continues and type your answer without any commas.

The population in 2016 will be
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Two linear data points ( time, population ).

(0,46020) and (12,52070).

Year 2016 corresponds to some point (16,y).

You can start by , and compute this.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
A town’s population has been increasing at a constant rate.
In 2010 the population was 46,020. By 2012 the population had increased to 52,070 people.
Assume this trend continues and type your answer without any commas.
The population in 2016 will be
~~~~~~~~~~~~~~~~~~

The problem does not specify, which the trend is: linear or exponential.

Also, it does not specify, if the rate is linear or exponential.

With such uncertainty, the problem can not be solved properly.

It requires making assumptions.

If the trend and the rate are linear, then there is no need to create interpolation function:

simply compute the increment from 2010 to 2012 and then add the doubled increment to the population of the year of 2012.

        increment = 52,070 - 46,020 = 6,050

        population in the year 2016 = 52,070 + 2*6,050 = 64,070.         ANSWER

If the trend is exponential, then the answer will be different.

It is not the tutors' job to guess what is a correct formulation and what is a hidden meaning of this problem.

---------------

Be aware, since the formula in the post by @josgarithmetic is incorrect,

and you will get wrong answer, if you try use it . . .

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