Question 626098
The cost, in millions of dollars, to remove x % of pollution in a lake is modeled by 6000+20(x)/300-3(x) .  Hint: Example, if x%=30%, then use x=30 in the C formula (do not convert x to a decimal)
a.	What is the cost to remove 65% of the pollutant?(show work) 
C= 6000+20(65)/300-3(65)
=7300/105
C=69.5
It would cost around 69.5 million dollars to remove 65% of the pollutant in the lake.

b.	What is the cost to remove 85% of the pollutant?(show work)
 C=6000+20(85)/300-3(85)
=7700/45
C=171.1
It would cost 171.1 million dollars to remove 85%of the pollutant in the lake.
c.	What is the cost to remove 98% of the pollutant?(show work)
 C=6000+20(98)/300-3(98)
=7960/6
C=1326 (rounded)
I(t would cost 1326 million dollars to remove 98% of the pollutant in the lake.
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Your answers are correct.
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d.For what value is this equation undefined?(show work)

It's undefined if the denominator = zero, which is removing 100%
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300 - 3x = 0
x = 100 --> undefined