According to data from the U.S. Census Bureau, the population P of Cleveland, Ohio (in thousands) in year x can be approximated by the equation
P = .08x^2 - 13.08x + 927 , where x = 0 corresponds to the year 1950. In what year (in the past) was the population about 804, 200?
Well, solving this equation is no joke! Let's set P to 804.2 (since P is in thousands) and solve for x:
804.2 = .08x^2 - 13.08x + 927
After doing some math, we find that x is approximately equal to 50.5.
Since x=0 corresponds to the year 1950, we can calculate the year by subtracting 50.5 from 1950. And the result is...
Hopefully, you have your calculator ready, because the answer is approximately the year 1899.5! Did you spot any clowns back then?
To find the year in which the population was about 804,200, we need to solve the equation:
804.2 = 0.08x^2 - 13.08x + 927
We can rearrange the equation to set it equal to zero:
0.08x^2 - 13.08x + 927 - 804.2 = 0
Now we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 0.08, b = -13.08, and c = 927 - 804.2.
Plugging in these values, we get:
x = (-(-13.08) ± √((-13.08)^2 - 4 * 0.08 * (927 - 804.2))) / (2 * 0.08)
Simplifying further:
x = (13.08 ± √(170.6464 - 60.448)) / 0.16
x = (13.08 ± √110.1984) / 0.16
Now let's calculate the values:
x = (13.08 + √110.1984) / 0.16 ≈ 87.334
or
x = (13.08 - √110.1984) / 0.16 ≈ 53.166
Since x represents the number of years after 1950, we need to subtract these values from 1950 to get the corresponding years:
For x ≈ 87.334:
Year ≈ 1950 - 87.334 ≈ 1862.666
For x ≈ 53.166:
Year ≈ 1950 - 53.166 ≈ 1896.834
Therefore, the population of Cleveland, Ohio was about 804,200 in the past around the year 1862 or 1897.
To find the year in which the population of Cleveland, Ohio was about 804,200, we can set up an equation with the given population value and solve for x.
The equation for the population is:
P = 0.08x^2 - 13.08x + 927
We want to find the year when the population was approximately 804,200, so we can substitute this value for P:
804.2 = 0.08x^2 - 13.08x + 927
Now, we can solve this quadratic equation to find the value of x:
0.08x^2 - 13.08x + 927 - 804.2 = 0
0.08x^2 - 13.08x + 122.8 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the values are:
a = 0.08
b = -13.08
c = 122.8
Substituting these values into the quadratic formula:
x = (-(-13.08) ± √((-13.08)^2 - 4(0.08)(122.8))) / (2(0.08))
Simplifying further:
x = (13.08 ± √(171.5664 - 39.104)) / 0.16
x = (13.08 ± √(132.4624)) / 0.16
x = (13.08 ± 11.5105) / 0.16
Now, we can calculate the two possible values of x:
x1 = (13.08 + 11.5105) / 0.16
x1 = 142.1937 / 0.16
x1 = 888.71
x2 = (13.08 - 11.5105) / 0.16
x2 = 1.5695 / 0.16
x2 = 9.8094
Since x = 0 corresponds to the year 1950, we need to subtract x1 and x2 from 1950 to find the years when the population was approximately 804,200:
Year1 = 1950 - 888.71 ≈ 1061.29
Year2 = 1950 - 9.8094 ≈ 1940.19
Therefore, the population of Cleveland, Ohio was approximately 804,200 in the years around 1061 and 1940.