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College Math 117 Week 4 Answer Guide- Algebra 1B -

Week Four Algebra 1B study guide and Soultions

 

1. The cost, in millions of dollars, to remove % of pollution in a lake modeled by

a. What is the cost to remove 75% of the pollutant?

It will cost $120 million to remove 75% of the pollutant.

b. What is the cost to remove 90% of the pollutant?

It will cost $150 million to remove 90% of the pollutant.

c. What is the cost to remove 99% of the pollutant?

It will cost $3,000 million or $3 billion dollars to remove 99% of the pollutant.

d. For what value is this equation undefined?

Set denominator = 0 and solve:

200 – 2x = 0

200 = 2x

100 = x

Equation is undefined for = 100%.

e. Do the answers to sections a. through d. match your expectations? Explain why or why not.

Answers may vary but must indicate that the cost to remove the pollutant increases as the amount of pollutant removed increases and that it is impossible to remove all of the pollutant from the lake.

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 animals.

Letting = number of animals tested and = total costs

C = Setup costs + cost to test

C = 2500 + 3x

c. Find the average cost per animal for 10 animals, 100 animals, and 1,000 animals.

Number of animals (x) Cost Simplified answer

10

10

2500 + 3*10 $253.00

100

100

2500 + 3*100 $28.00

1000

1000

2500 + 3*1000 $5.50

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.

As the number of animals tested increases, the average cost per animal decreases. The average cost per animal will never drop below $3.00 per animal, though. The average cost decreases, because, in the average, the $2,500 setup cost is distributed evenly among the number of animals, so testing many animals makes the setup cost per animal very small.

e. How many animals should be tested for the average cost to be $5.00 per animal?

1,250 animals must be tested for the average cost to drop to $5.00 per animal.

3. To estimate animal populations, biologists count the total number of animals in a small section of the 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.

Letting = number of animals, and = amount of land,

A = kL

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.

A = kL

12 = k*100

k = 12/100 = 0.12

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.

Using the answer from part b., we know that

= 0.12 L

We also know that L = 2500, so

A = 0.12*2500

A = 300

We would expect there to be 300 white tail deer in the entire preserve.

 

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