Let's address the given problem step-by-step:
### Part a: Compute the cost to remove 25% of the air pollutants.
To find the cost to remove 25% of the air pollutants, we substitute [tex]\( x = 25 \)[/tex] into the cost function [tex]\( C(x) \)[/tex]:
[tex]\[
C(25) = \frac{500 \times 25}{140 - 25}
\][/tex]
Simplifying the denominator, we get:
[tex]\[
140 - 25 = 115
\][/tex]
Now, substituting back:
[tex]\[
C(25) = \frac{500 \times 25}{115} = \frac{12500}{115}
\][/tex]
Simplifying the fraction:
[tex]\[
C(25) = 108.69565217391305 \, \text{(in thousands of dollars)}
\][/tex]
Since the cost is given in thousands of dollars, we convert this value to actual dollars:
[tex]\[
Cost \ to \ remove \ 25\% = 108.69565217391305 \times 1000
= 108,695.65217391304 \, \text{dollars}
\][/tex]
So the cost to remove 25% of the air pollutants is [tex]\( \$108,695.65 \)[/tex].
### Part b: If the power company budgets \[tex]$1.4 million for pollution control, what percentage of the air pollutants can be removed?
Given the budget, \( \$[/tex]1.4 \) million, which is equivalent to [tex]\( 1.4 \times 1000 = 1400 \)[/tex] thousand dollars. We need to find [tex]\( x \)[/tex] such that:
[tex]\[
\frac{500 x}{140 - x} = 1400
\][/tex]
To solve for [tex]\( x \)[/tex], we first set up the equation:
[tex]\[
1400 \times (140 - x) = 500 x
\][/tex]
Expansion and rearrangement gives:
[tex]\[
196000 - 1400 x = 500 x
\][/tex]
Combining like terms:
[tex]\[
196000 = 500x + 1400x
\][/tex]
[tex]\[
196000 = 1900x
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{196000}{1900}
= 103.15789473684211 \%
\][/tex]
Thus, if the power company budgets \[tex]$1.4 million for pollution control, they can remove approximately \( 103.16 \% \) of the air pollutants.
### Conclusion:
Therefore, the correct answers are:
- a. The cost to remove 25% of the air pollutants is \(\frac{2500}{23}\).
- b. The percentage of air pollutants that can be removed with a $[/tex]1.4 million budget is [tex]\( 103.16 \% \)[/tex].
Thus, the selection from the given options is:
d. a. [tex]\(\frac{2500}{23}\)[/tex], b. [tex]\(103.16 \%\)[/tex]
