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The population of a town grows exponentially. After 1 year, the population is 34,560. After 2 years, the population is 37,325. Provide an equation that can be used to predict, y, the number of people living in the town after x years. (Round population values to the nearest whole number.)

Respuesta :

carlosego

For this case we have a function of the form:
 [tex]y = A * (b) ^ x [/tex]
 Where,
 A: initial amount
 b: growth rate
 x: time in years
 We must find the values of A and b.
 For this, we use the following data:
 After 1 year, the population is 34,560:
 [tex]A * (b) ^ 1 = 34560 [/tex]
 After 2 years, the population is 37,325:
 [tex]A * (b) ^ 2 = 37325 [/tex]
 We have a system of two equations with two unknowns.
 By dividing both equations we obtain the value of b:
 [tex] \frac{A * (b) ^ 2}{A * (b) ^ 1} = \frac{37325}{34560} [/tex]
 [tex]b = \frac{37325}{34560} [/tex]
 [tex]b = 1.08[/tex]
 Then, the value of A is obtained by substituting b in any of the equations.
 We have then:
 [tex]A * (b) ^ 1 = 34560 A * (1.08) ^ 1 = 34560 A * (1.08)= 34560 [/tex]
 [tex]A = \frac{34560}{1.08} [/tex]
 [tex]A = 32000[/tex]
 Answer:
 an equation that can be used to predict, y, the number of people living in the town after x years is:
 [tex]y=32000*(1.08)^x[/tex]

Answer:

Equation will be [tex]y=(32083).(1.08)^{x}[/tex]

Step-by-step explanation:

Since population of a town grows exponentially so the function to represent the population growth will be

[tex]a_{n}=a_{0}(r)^{n}[/tex]

n = number of years

where [tex]a_{n}[/tex] = population after n years

[tex]a_{0}[/tex] = population in the base year or population at 0 year

r = common ratio or the growth with which population is increasing year by year

Now It is given in the question after 1 year, the population is 34560

So the expression will be [tex]34560=a_{0}(r)^{1}=a_{0}.(r)[/tex]------(1)

again question states that after 2 years, the population is 37325

The expression becomes [tex]a_{2}=37325=a_{0}.(r)^{2}[/tex]------(2)

Now we divide equation 2 from equation 1

[tex]\frac{37325}{34560}=\frac{a_{0}.r^{2}}{a_{0}.r}[/tex]

r = 1.08

Now we put the value of "r" in equation 1 to calculate the value of [tex]a_{0}[/tex]

[tex]34650=a_{0}.(1.08)[/tex]

[tex]a_{0}= \frac{34650}{1.08}=32083[/tex]

If y be the number of people living in the town after x years then the function formed will be

[tex]y=(32083).(1.08)^{x}[/tex]

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