SOLUTION: Can someone please help: Consider this scenario: A town's population has been decreasing at a constant rate. In 2010 the population was 5,700. By 2013 the population had dropped

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Question 1135854: Can someone please help:
Consider this scenario: A town's population has been decreasing at a constant rate. In 2010 the population was 5,700. By 2013 the population had dropped 5,400. Assume this trend continues.
Predict the population in 2016.

Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39625)   (Show Source): You can put this solution on YOUR website!
"decreasing at a constant rate" suggests linear relationship.

Your description for year 2013 said, "dropped 5400". That means, the population in year 2013 was . Description did not say "dropped TO 5400".

According to the description as posted,
points (x,y) for x being year past 2010 and y being population
(0, 5700) and (13, 300).

Equation for the line:



which is probably not the actual equation you want, but the method should be clear and you can fix the possible mistake in your description, and find the actual equation for your corrected description.

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
in 2010 the population was 5700.
in 2013 the population was 5400.

that's a loss of 300 in 3 years which becomes an average loss of 300 / 3 = 100 per year.

if this trend continues, the population in 2016 is predicted to be 5400 - (2016 - 2013) * 100 = 5400 - 3 * 100 = 5400 - 300 = 5100.

this assumes a straight line loss which is a constant amount per year.

the equation for the straight line loss would be y = 5700 - 100 * x, where x is the number of years since 2010.

this can be graphed and looks like this.

$$$

when x = 3, the year is 2010 + 3 = 2013.
when x = 6,the year is 2010 + 6 = 2016.

the above formuls is used if the drop in the population is at a constant amount per year.

if the drop in the population is at a constant rate per year, then the formula is different.

to calculate the average rate of loss, use the following formula.

f = p * (1 + r) ^ n

f is the future value
p is the present value
r is the interest rate per time period (assumed to be years in this problem).
n is the number of time periods (assumed to be years in this problem.

since p = 5700 and f = 5400, and n = 3 (from 2010 to 2013), the formula becomes:

5400 = 5700 * (1 + r) ^ 3

divide both sides of this equation by 5700 to get 5400 / 5700 = (1 + r) ^ 3

take the third root of both sides of this equation to get (5400 / 5700) ^ (1/3) = 1 + r.

subtract 1 from both sides of this equation to get (5400 / 5700) ^ (1/3) - 1 = r

solve for r to get r = -.0178609748.

that's your average interest rate per year.

confirm by using that rate in the formula of f = p * (1 + r) ^ 3

when p = 5700 and r =-.0178609748, the formula becomes f = 5700 * (1 - .0178609748) ^ 3 = 5400, which is correct, so the interest rate per year is good.

to find the population in s016, use the formula of f = p * (1 + r) ^ n, where:

p = 5700
r =-.0178609748
n = (2016 - 2010) = 6

to get f = 5700 * (1 - .0178609748) ^ 6 = 5115.789474 which rounds to 5116.

this formula can be graphed and looks like this.

$$$

the difference between the methods is not so pronounced in only 6 years.

if you go out more years, the difference will be more pronounced.

going out 20 years, or even 30 years, you can see the difference in the following graph that shows both methods on the same graph.

$$$

the red line is the constant amount per year loss.
formula is y = 5700 - 100 * x.

the blue line is the constant rate per year loss.
formula is y = 5700 * (1 - .0178609748) ^ x.

the values for the values of x indicated are:



year     x    red line     blue line    blue line minus red line


2010     0    5700         5700         0


2013     3    5400         5400         0


2016     6    5100         5116         16


2030     20   3700         3975         275


2040     30   2700         3319         619













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