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Sagot :
The population is growing at the rate of 2% per year.
So in 'y' years from the start, its size is modeled by the function
Population = (original size) x (1.02)^y power
This equation is familiar from all of the bank-interest problems.
9). The population is 765 right now. In 'y' more years from now, it will be
Population = 765 x (1.02)^y power
11). In 5 years from now, it will be P = 765 (1.02)^5 = 845 students (rounded)
12). The population will exceed 1,000 students when
1,001 = 765 (1.02)^y
Divide each side by 765 : 1,001/765 = (1.02)^y
Take the log of each side: log(1,001/765) = y log(1.02)
Divide each side by log(1.02) : y = log(1,001/765) / log(1.02)
Fire up your calculator and crank out that number:
log(1,001/765) / (log(1.02) = 0.11677 / 0.0086 = 13.5779 years
If the 2% growth is steady and uniform 24/7 throughout the year, then
this number means that the 1,001th student will enroll in
13 years 30 weeks 1 day 2 hours 6 minutes 36 seconds from right now.
So in 'y' years from the start, its size is modeled by the function
Population = (original size) x (1.02)^y power
This equation is familiar from all of the bank-interest problems.
9). The population is 765 right now. In 'y' more years from now, it will be
Population = 765 x (1.02)^y power
11). In 5 years from now, it will be P = 765 (1.02)^5 = 845 students (rounded)
12). The population will exceed 1,000 students when
1,001 = 765 (1.02)^y
Divide each side by 765 : 1,001/765 = (1.02)^y
Take the log of each side: log(1,001/765) = y log(1.02)
Divide each side by log(1.02) : y = log(1,001/765) / log(1.02)
Fire up your calculator and crank out that number:
log(1,001/765) / (log(1.02) = 0.11677 / 0.0086 = 13.5779 years
If the 2% growth is steady and uniform 24/7 throughout the year, then
this number means that the 1,001th student will enroll in
13 years 30 weeks 1 day 2 hours 6 minutes 36 seconds from right now.
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