Axia College Material Appendix D The Environment Environmental

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Axia College Material
Appendix D
The Environment
Environmental and wildlife preservation ensures future generations will have available resources and can
enjoy the beauty of Earth. On the other hand, some of the damage humans have already done to the
environment is expensive and time consuming to eradicate.
Application Practice
Answer the following questions. Use Equation Editor to write mathematical expressions and equations.
First, save this file to your hard drive by selecting Save As from the File menu. Click the white space
below each question to maintain proper formatting.
1. The cost, in millions of dollars, to remove x % of pollution in a lake modeled by C 
6,000
200  2 x
a. What is the cost to remove 75% of the pollutant?
b. What is the cost to remove 90% of the pollutant?
c.
What is the cost to remove 99% of the pollutant?
d. For what value is this equation undefined?
e. Do the answers to sections a. through d. match your expectations? Explain why or why
not.
Busch Enteratinment Corporation. (N.D.). Flamingos. Retrieved from
http://www.seaworld.org/animal-info/info-books/flamingo/physical-characteristics.htm
Answers
a) If x= 75 then
C
6,000
6,000

 120
200  2(75)
50
Answer: The cost to remove 75% of the pollutant is 120 millions of dollars
b) If x= 90 then
C
6,000
6,000

 300
200  2(90)
20
Answer: The cost to remove 90% of the pollutant is 300 millions of dollars
c)
If x= 95 then
C
6,000
6,000

 3,000
200  2(99)
2
Answer: The cost to remove 99% of the pollutant is 3,000 millions of dollars
MAT/117
d) If x=100 then
C
6,000
6,000
is undefined

200  2(100)
0
(100 does not belong to the domain of C)
Answer: for x=100 equation is undefined
e) Yes,from a) to e) the cost to remove x% of the pollution is increasing because x is
incerasing. This fact is reasonable because the cost increases if we want a higher %
to be removed
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.
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.
c.
Find the average cost per animal for 10 animals, 100 animals, and 1,000 animals.
d. As the number of animals tested increases, what happens to the average cost to test the
animals? Would the average cost ever fall below $3.00? If so, identify a value that
supports your answer. If not, explain how you know.
e. How many animals should be tested for the average cost to be $5.00 per animal?
C  3 x  2,500
3x  2,500
b) AC 
x
3(10)  2,500 2,530

 253
c) If x=10 AC 
10
10
a)
Answer: $253
If x=100 AC 
3(100)  2,500 2,800

 28
100
100
Answer: $28
If x=1000 AC 
3(1000)  2,500 5,500

 5.5
1000
1000
Answer: $5.5
d) AC 
3 x  2,500
2,500
 3
, x>0 then AC>3 always
x
x
Answer:
Average cost decreases as the number of animals increase and never is below 3
3x  2,500
 5  3x  2,500  5 x
e)
x
 3x  5 x  2,500  2 x  2,500  x  1,250
AC 
MAT/117
Answer: 1,250 animals
3. To estimate animal populations, biologists count the total number of animals in a small section of
a habitat. The total population of animals is directly proportional to the size of the habitat (in
acres) polled.
a. Write an equation using only one variable that could be used to solve for the constant of
variation k.
Let a=size of habitat in acres
N= number of animals in the habitat
Equation: N=k*a
b. A biologist counted 12 white tail deer in a 100-acre parcel of land in a nature preserve.
Find the constant of variation k.
12=k*100 then k= 12/100= 3/25
Answer: k=3/25
c.
If the entire nature preserve is 2,500 acres, then what is the total white tail deer
population in the preserve? Describe how you arrived at your answer.
If we have 12 animals in 100 acres then we have x animals in 2,500 acres
where 12/100 = x/2,500 then x= 12*2,500/100=300
Answer: 300 animals
MAT/117
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