We presume your cost function is
c(p) = 124p/((10 +p)(100 -p))
This can be rewritten as
c(p) = (124/11)*(10/(100 -p) -1/(10 +p))
The average value of this function over the interval [50, 55] is given by the integral

This evaluates to
(-124/55)*(ln(65/60)+10ln(45/50)) ≈ 2.19494
The average cost of removal of 50-55% of pollutants is about
$2.19 hundred thousand = $219,000
Answer:
The population of bacteria can be expressed as a function of number of days.
Population =
where n is the number of days since the beginning.
Step-by-step explanation:
Number of bacteria on the first day=![\[5 * 2^{0} = 5\]](https://tex.z-dn.net/?f=%5C%5B5%20%2A%202%5E%7B0%7D%20%3D%205%5C%5D)
Number of bacteria on the second day = ![\[5 * 2^{1} = 10\]](https://tex.z-dn.net/?f=%5C%5B5%20%2A%202%5E%7B1%7D%20%3D%2010%5C%5D)
Number of bacteria on the third day = ![\[5*2^{2} = 20\]](https://tex.z-dn.net/?f=%5C%5B5%2A2%5E%7B2%7D%20%3D%2020%5C%5D)
Number of bacteria on the fourth day = ![\[5*2^{3} = 40\]](https://tex.z-dn.net/?f=%5C%5B5%2A2%5E%7B3%7D%20%3D%2040%5C%5D)
As we can see , the number of bacteria on any given day is a function of the number of days n.
This expression can be expressed generally as
where n is the number of days since the beginning.
Answer:
Step-by-step explanation:
<h3>Given</h3>
- f(n) = n ^ 4 + 5n ^ 3 + 2n ^ 2 - 5n + 13
<h3>To find </h3>
<h3>Solution</h3>
<u>Substituting n with -2</u>
- f(-2) =
- (-2)^4 + 5(-2)^3 + 2(-2)^2 - 5(-2) + 13 =
- 16 - 40 + 8 + 10 + 13 =
- 7