The next three problems explore some techniques of data analysis. 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 b=__________. Suppose you want to estimate the population in 1975. To that end you compute N(1975) = _________. (Round your answers to the nearest integer. The process illustrated in this problem is called linear interpolation.)
To find the values of m and b, we can use the given information that N(1950) = A and N(2000) = B.
First, let's find the slope, m:
m = (B - A) / (2000 - 1950)
m = (6079006982 - 2555360972) / (2000 - 1950)
m = 3523646010 / 50
m = 70472920.2
Next, let's find the y-intercept, b:
We know that N(1950) = A, so we can substitute these values into the equation:
A = m * 1950 + b
2555360972 = (70472920.2 * 1950) + b
2555360972 = 137378277790 + b
b = 2555360972 - 137378277790
b = -134822917818
So, m = 70472920.2 and b = -134822917818.
Now, let's find the estimated population in 1975 using linear interpolation:
N(1975) = m * 1975 + b
N(1975) = (70472920.2 * 1975) + (-134822917818)
N(1975) = 139199115975 - 134822917818
N(1975) = 4371194157
Therefore, the estimated population in 1975 is 4371194157.
To find the values of m and b in the equation N(t) = mt + b, we need to solve the system of equations using the given information:
1) N(1950) = A:
A = m(1950) + b
2) N(2000) = B:
B = m(2000) + b
Substituting the given values of A and B into the equations, we have:
1) A = 2555360972m + b
2) B = 6079006982m + b
We can now solve this system of equations. Subtract equation 1 from equation 2 to eliminate b:
(B - A) = (6079006982m + b) - (2555360972m + b)
B - A = 6079006982m - 2555360972m
B - A = 3523646010m
Solving for m:
m = (B - A) / 3523646010
Now, substitute the value of m back into equation 1 to solve for b:
A = 2555360972 * [(B - A) / 3523646010] + b
A - 2555360972 * [(B - A) / 3523646010] = b
Substitute the given values of A and B to find b:
b = A - 2555360972 * [(B - A) / 3523646010]
Calculating the values:
m = (6079006982 - 2555360972) / 3523646010 ≈ 1.3744
b = 2555360972 - 2555360972 * [(6079006982 - 2555360972) / 3523646010] ≈ -1378205423
To estimate the population in 1975, substitute t = 1975 into the equation N(t) = mt + b:
N(1975) ≈ 1.3744 * 1975 - 1378205423 ≈ 3864
Therefore, the estimated population in 1975 is 3864 (rounded to the nearest integer).
To find the values of m and b in the equation N(t) = mt + b using the given information, we can set up a system of two equations with the known values for t and N.
From the problem statement:
N(1950) = A, and N(2000) = B
Plugging these values into the equation, we get:
A = m(1950) + b ----(1)
B = m(2000) + b ----(2)
To solve this system of equations, we can subtract equation (1) from equation (2):
B - A = m(2000) + b - (m(1950) + b)
B - A = m(2000) - m(1950)
Since b cancels out, we have:
B - A = m(2000 - 1950)
B - A = 50m
Now we can solve for m:
m = (B - A) / 50
We can substitute the given values of A and B into the equation to find m:
m = (6079006982 - 2555360972) / 50
m ≈ 70672820
Now that we have the value of m, we can substitute it into equation (1) to find b:
A = (70672820)(1950) + b
b = A - (70672820)(1950)
Now we can calculate the value of b:
b ≈ 2562194612
Therefore, the values of m and b are approximately m = 70672820 and b = 2562194612.
Now, to estimate the population in 1975, we can use the linear interpolation technique. We substitute t = 1975 into the equation N(t) = mt + b:
N(1975) = (70672820)(1975) + 2562194612
Calculating this expression, we get:
N(1975) ≈ 4392262702
Therefore, the estimated population in 1975 is approximately 4,392,262,702.