Posted by Bretagne on Thursday, January 31, 2008 at 3:01pm.
1a. C = 6000/(200-2x), with x in %
For x = 90%, that is 6000/20 = 300 (million $)
You must be doing something wrong.
2a. No
C = 2500 + 2x
You say the fixed cost is $2500. Where did the $3500 figure come from?
See how you do on the rest of the questions after making those corrections
a.What is the cost to remove 75% of the pollutant?
Answer: $120 million dollars
b.What is the cost to remove 90 % of the pollutant?
Answer: $300 million dollars
c.What is the cost to remove 99% of the pollutant?
Answer: $3 billion dollars
d.For what value is this equation undefined?
I don't understand this
2.Biologists want to set up a station to test alligators in the lake for West Nile Virus. Suppose that the costs for such a station are $2,500 for setup costs and $3.00 to administer each test.
a.Write an expression that gives the total cost to test x animals.
I meant to write 2500 and I got the 3 from above. The $3.00.
2500 + 3x
b.You can find the average cost per animal by dividing total costs by number of animals. Write the expression that gives the average cost per animal.
2500+3x/x
THIS IS YOUR HOMEWORK FROM A COLLEGE LEVEL COURSE! DO IT YOURSELF!
Eat it.
Worse than just homework from a college level course, it is two identical questions from one of the exams;I certainly hope you just used the formula and figured it out for yourself.
However, your question did help me to realize that I was converting 75% to .75 which was not giving me the right answer - apparently it needs to be left as 75. Once I realized this, I did the work - including the first problem - on my own and checked it. If you did the same, ok.
But when you ask for help with an exam question and do not substitute different values than what is on the test, I believe it qualifies as cheating. You might want to keep an eye on that because if you get caught they might drop you from the school.
I think it is nothing wrong for asking for homework help. I am sure you have ask for help in school before so stop trying to be hateful and how do you know if he or she is asking a test question. If you do not want to help then go away and as for the person who was asking for help I was trying to figure out if I should convert 75 into a decimal to, so I understand as well now. Good luck with homework for the person who is asking for help.
This is an exact copy of Appendix D from week 4 of MAT/117 at University of Phoenix. (Not an exam. But an entire homework assignment.) Every single assignment can be found posted to many different sites by many different students. I sometimes look them up to check a few of my answers. ^__^
You know what I find funny? The fact that the people claiming those asking for help are cheaters. Question I have to ask is what are you doing here?
Next, as long as your using this site for actual help and not copying everything then I think its fine.Need help setting up an equation? Just fine. Looking for all the right answers? Your cheating.
Hey, If people want to post the answer and you smart enough to get it only you have to pay the price, other than that who cares if one is cheating.