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19.10.2019 •
Mathematics
The population of mastertown was 23,000 in 2012. assume that mastertown's population increases at a rate of 2% per year. write an equation to model the population of mastertown (y) based on number of years since 2012 (x).
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Ответ:
The population was 23,000 in 2012→y(0)=23,000
The population increases at a rate of 2% per year.
If the population of Mastertown is y; and the number of years since 2012 is x, then:
In year 2013, x=2013-2012→x=1 (1 year since 2012), the population would be:
y(1)=y(0)+2% y(0)→y(1)=23,000+2%(23,000)
Getting 23,000 common factor:
y(1)=23,000 (1+2%)→y(1)=23,000 (1+2/100)→y(1)=23,000 (1+0.02)→y(1)=23,000 (1.02)
In year 2014→x=2014-2012→x=2 (2 years since 2012), the population would be:
y(2)=y(1)+2% y(1)=23,000(1.02)+2%[23,000(1.02)]
Getting common factor 23,000(1.02):
y(2)=23,000(1.02)(1+2%)→y(2)=23,000(1.02)(1+2/100)→y(2)=23,000(1.02)(1+0.02)→
y(2)=23,000(1.02)(1.02)→y(2)=23,000(1.02)^2
In year 2015→x=2015-2012→x=3 (3 years since 2012), the population would be:
y(3)=y(2)+2% y(2)=23,000(1.02)^2+2%[23,000(1.02)^2]
Getting common factor 23,000(1.02)^2:
y(3)=23,000(1.02)^2 (1+2%)→y(3)=23,000(1.02)^2 (1+2/100)→y(3)=23,000(1.02)^2 (1+0.02)→
y(3)=23,000(1.02)^2 (1.02)→y(3)=23,000(1.02)^3
Then for x years from 2012, the population in Mastertown would be:
y(x)=23,000(1.02)^x
The equation to model the population of Mastertownn is y=23,000(1.02)^x
Ответ:
The equation for the growth of the population is given by:
where
a is the initial amount
n represents the number of years and
r represents the growth rate( in decimal)
As per the statement:
The population of Master-town was 23,000 in 2012
Here, y represents the population of Master-town (y) based on number of years since 2012 (x).
⇒ a= 23000
It is also given that Master-town's population increases at a rate of 2% per year.
⇒![r = 2\% = 0.02](/tpl/images/0334/7260/f00dc.png)
then;
Therefore, an equation to model the population of Master-town (y) based on number of years since 2012 (x) is:
Ответ: