Assume that the population of the world in 2010 was 6.9billion and is growing at the rate of 1.1% a year. a) Set up a recurrence relation for the population of theworld n years after 2010. b) Find an explicit formula for the population of theworld n years after 2010. c) What will the population of the world be in 2030? 21. A factory makes custom sports cars at an increasing rate

Answer:a) If we call P(n) the population n years after 2010, the recurrence relation for the population of the world n years after 2010 would beP(0) = 6.9 billionP(n) = P(n-1)*(1.011)b) [tex]P(n)={6.9*(1.011)^n[/tex]c)[tex]6.9*(1.011)^{20} = 8.5876\; billion[/tex]Step-by-step explanation:a)If the growing rate is 1.1% in the year 2011 was6.9 + 1.1% of 6.9 = 6.9 + 6.9*(0.011) = 6.9*(1.011)In the year 2012, the new population was6.9*(1.011) + 1.1% of 6.9*(1.011)= 6.9*(1.011) + 6.9*(1.011)*(0.011) = 6.9*(1.011)*(1+0.011)= 6.9*(1.011)*(1.011) = [tex]6.9*(1.011)^2[/tex]Similarly, we can see that the population in 2013 was[tex]6.9*(1.011)^3[/tex]If we call P(n) the population n years after 2010, the recurrence relation for the population of theworld n years after 2010 would beP(0) = 6.9 billionP(n) = P(n-1)*(1.011)b)In the year n after 2010, the population would be[tex]P(n)={6.9*(1.011)^n[/tex]c)The population of the world in 2030, according to the formula, will be P(20)[tex]\boxed{P(20) = 6.9*(1.011)^{20} = 8.5876\; billion}[/tex]