Since january 1980 the population of the city of brownville has grown according to the mathematical model y=720,500(1.022)^x, where x is the number of years since january 1980.
explain what the numbers 720,500 and 1.022 represent in the model.
If this trend continues, use the model to predict the year during which the population of brownville will reach 1,548,800.
2007-12-30 17:31:38 · 5 answers · asked by Lovelygirl 1 in Science & Mathematics ➔ Mathematics
720,500 is initial population
1.022 is growth factor
15488000 = 7205000(1.022)^x
1.022^x = 2.15
x ln 1.022 = ln 2.15
x = ln 2.15 / ln 1.022
x = 35
Answer is 1980 + 35 = 2015
2007-12-30 20:28:32 · answer #1 · answered by Como 7 · 3⤊ 0⤋
720,500 and 1.022 are constants arrived at by conducting regression analysis of historical data.
1548800= 720500 (1.022)^x
1.022^x = 1548800/720500= 2.14962
x ln(1.022) = ln (2.14962)
0.02176x = .76529
x= 35.17 or 1980 + 35.17 = 2015 March 2
2007-12-31 01:51:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
1,548,800 = 720,500(1.022)^x
1,548,800/720,500 = 1.022^x
log(base1.022) (1,548,800/720,500) = x
x = ln(1,548,800/720,500)/ln(1.022) = 35.167180116433115903599100116675
Thus the year in which brownvilles population will reach 1,548,800 is 1980+35 = 2015.
2007-12-31 01:46:34 · answer #3 · answered by Peter R 1 · 0⤊ 0⤋
720,500 is population in year 1980.
22 per thousand is the rate of increase per year.
1, 548,800 will be after 35 years in 2015 given by
720,500(1.022)^x = 1,548,800
whence (1.022)^x = 2.1494 giving
x = log(2.1496) / log(1.022) = 34. 9611
2007-12-31 01:52:02 · answer #4 · answered by sv 7 · 0⤊ 0⤋
y=720,500(1.022)^x
1 548 800 = 720 500 (1.022)^x
1 548 800/720 500 = 1.022^x
log(1 548 800/720 500) = xlog 1.022
x = log(1 548 800/720 500)/log 1.022
x = 35 years approximately
1980 + 35 = 2015 ANS
teddy boy
2007-12-31 01:48:34 · answer #5 · answered by teddy boy 6 · 0⤊ 0⤋