I want you to do a linear and exponential model for Yakima's Population. Use the DATA below: in 1990 => P0 = 64792 in 2010 => P20 = 91646 A) Find the linear model of population growth. B) Predict the population for 2050. C) Find the exponential model of population growth. D) Predict the population for 2050. E) Which one seems more realistic?

Answer:

(a) [tex]P(x) = 1342.7x + 64792[/tex]

(b) [tex]P(60) = 145354[/tex]

(c) [tex]P(x) = 64792 *1.0175^x[/tex]

(d) [tex]P(60) = 183491[/tex]

(e) The exponential model is more realistic

Step-by-step explanation:

Given

[tex]1990 ==> P_0 =64792[/tex]

[tex]2010 ==> P_{20} =91646[/tex]

Solving (a): The linear model of the population growth

First, we calculate the slope of the function.

[tex]m = \frac{P_{20} - P_0}{20 - 0}[/tex]

[tex]m = \frac{91646 - 64792}{20 - 0}[/tex]

[tex]m = \frac{26854}{20}[/tex]

[tex]m = 1342.7[/tex]

The linear equation is then calculated using:

[tex]P(x) - P_0 = m(x - x_0)[/tex]

[tex]P(x) - 64792 = 1342.7(x - 0)[/tex]

[tex]P(x) - 64792 = 1342.7(x)[/tex]

[tex]P(x) - 64792 = 1342.7x[/tex]

Make P(x) the subject

[tex]P(x) = 1342.7x + 64792[/tex]

Solving (b): The population in 2050

First, calculate x:

[tex]x = 2050 - 1990[/tex]

[tex]x = 60[/tex]

Substitute 60 for x in [tex]P(x) = 1342.7x + 64792[/tex]

[tex]P(60) = 1342.7 * 60 + 64792[/tex]

[tex]P(60) = 80562 + 64792[/tex]

[tex]P(60) = 145354[/tex]

The population in 2050 is 145354

Solving (c): The exponential model of the population growth

An exponential model is:

[tex]y = ab^x[/tex]

In this case, it is:

[tex]P(x) = P_0 * b^x[/tex]

For x = 20, we have:

[tex]P(20) = P_0 * b^{20[/tex]

Substitute values for P(20) and P0

[tex]91646 = 64792 * b^{20[/tex]

Divide both sides by 64792

[tex]\frac{91646 }{64792} = \frac{64792 * b^{20}}{64792}[/tex]

[tex]\frac{91646 }{64792} = b^{20}[/tex]

[tex]1.41446474873 = b^{20}[/tex]

Take the 20th root of both sides

[tex]\sqrt[20]{1.41446474873} = b[/tex]

[tex]b = \sqrt[20]{1.41446474873}[/tex]

[tex]b = 1.0175[/tex]

So, the model is:

[tex]P(x) = P_0 * b^x[/tex]

[tex]P(x) = 64792 *1.0175^x[/tex]

Solving (d): The population in 2050

First, calculate x:

[tex]x = 2050 - 1990[/tex]

[tex]x = 60[/tex]

Substitute 60 for x in [tex]P(x) = 64792 *1.0175^x[/tex]

[tex]P(60) = 64792 *1.0175^{60[/tex]

[tex]P(60) = 64792 *2.832[/tex]

[tex]P(60) = 183490.944[/tex]

[tex]P(60) = 183491[/tex] ---- approximated

(e) The most realistic model

The exponential model is more realistic. This is so because:

The linear model grows at a constant linear rate which means that, every year a certain amount of individuals is added to the society. However, this is not always so because it is almost impossible to for growth rate to be constant

A curve in the exponential model shows that the addition of individuals in the society every year is not always constant.

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