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October 1, 2014

Homework Help: math

Posted by lauren on Monday, August 28, 2006 at 5:10pm.

Thank goodness for this site! Two questions:

first: in the 1990 U.S. census, the final count was 248,709,873 people. The general population is estimated to have been undercounted by 1.8%, that is, only 98.2% of the people were counted.
If this were the case, what was the actual 1990 population?
Second:
Suppose f(x)=x^2-x-12
Solve f(x)=8
(the first question to this was evaluate f(8), so i substituted 8 in place of x and solved. I do not know what to do for f(x)=8



thank you so much!!!
Lauren


Ok, if .98*True_Population=248,709,873 then what is True_Population?
(hint: divide both sides by some decimal.)
___________
Now we have f(x)=x^2-x-12
Thus 8=x^2-x-12 , or x^2-x-20=0
Now solve for x by factoring or by the quadratic formula.



Looks like I dropped a number for the first one. it should be .982*True_Population=248,709,873


i understand the second, but the first is still confusing. you say both sides, but what is the equation?

.982x (x being the true population)= 248,709,873

or .018x=248,709,873


The first one.
We're told that the population has been understated. Therefore the number we're given, 248,709,873 , is .982 (or 98.2%) of the true population -or the more correct answer, since there is no such thing as the "exact" population when we're talking about numbers this size. There are births and deaths while the census is being taken. Thus,
.982x (x being the true population)= 248,709,873
would be correct here. Now if you divide both sides by .982 -just regular algebra- you should get the correct answer.

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