Question 418522
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Hi
1. The cost, in millions of dollars, to remove x % of pollution in a lake modeled by 
{{{C = 6000/(200-2x)}}} {{{6000/(200-2*75) = 120}}} {{{6000/(200-2*90)=300}}} {{{6000/(200-2*99)= 3000}}}
Equation undefined for x = 100%
Makes sense in that the more pollution removed, the higher the cost

2. C = $3x+ $2500
C[Average] = {{{$3 + $2500/x}}}  {{{3 + 2500/10 = 253}}} {{{3+2500/100 = 28}}} {{{3 + 2500/1000 = 5.50}}} 
Cost of testing/animal decreases as the number of animals tested increases.
 the cost/animal can never be less than $3 as $2500/x will never be a negative number
e.  5.00 = 3 + 2500/x  OR 2500/x = 2  x = 2500/2 = 1250 animals

3. P = kA  12 = k*100  k = 12/100 = .12
   P = .12*2500 = 300 white tail deer