I believe you posted this question awhile ago. Anyway, you should get that
N(1975) = 4,317,183,977
How?
Use the equation for linear interpolation:
y = y₀ + (y₁ - y₀) [ (x - x₀)/(x₁ - x₀) ]
Putting it into context...
N(1750) = A + (B - A)[(1975 - 1950)/(2000 - 1950)]
According to the US Bureau of the Census, the world population in the year 1950 was A=2555360972, and in 2000 it was B=6079006982. We'll use A and B so we don't have to keep writing those large and idiosyncratic numbers. We usually use y and x in the equation of a line, but in this and the following problem let's use N and t instead. t stands for time and N for the size of the population.
If
N(t)=mt+b
such that N(1950)=A and N(2000)=B, then m= and and b= .
3 answers
Whoops, in the last line, I meant " N(1975) " and not " N(1750) "
Much Ado About Nothing.
two ordered pairs (1950 , 2555360972) and
(2000 , 6079006982)
slope = (6079006982 - 2555360972)/(2000-1950)
= 70472920.2 ---> m for N(t)= mt+b
using the first point:
2555360972 = 70472920.2(1950) + b
b = -1.348668 x 10^11
so we have N(t) = 70472920.2t -1.348668 x 10^11
checking for 2nd point, if t = 2000
N(2000) = 70472920.2(2000) - 1.348668 x 10^11
= 6079007000 , close enough
two ordered pairs (1950 , 2555360972) and
(2000 , 6079006982)
slope = (6079006982 - 2555360972)/(2000-1950)
= 70472920.2 ---> m for N(t)= mt+b
using the first point:
2555360972 = 70472920.2(1950) + b
b = -1.348668 x 10^11
so we have N(t) = 70472920.2t -1.348668 x 10^11
checking for 2nd point, if t = 2000
N(2000) = 70472920.2(2000) - 1.348668 x 10^11
= 6079007000 , close enough