19) The rational function 19) \[ C(x)=\frac{130 x}{100-x}, 0 \leq x<100 \] describes the cost, $\mathrm{C}$, in millions of dollars, to inoculate $\mathrm{x} \%$ of the population against a particular strain of the flu. Determine the difference in cost between inoculating $85 \%$ of the population and inoculating $45 \%$ of the population. (Round to the nearest tenth, if necessary.)
Solution
Step 1 ：The problem is asking for the difference in cost between inoculating 85% of the population and inoculating 45% of the population. This means we need to calculate the cost for each percentage and then subtract the two costs. We can use the given function \(C(x)=\frac{130 x}{100-x}\) to calculate the cost for each percentage.
Step 2 ：First, we calculate the cost of inoculating 85% of the population. Substituting \(x = 85\) into the function, we get \(C(85) = \frac{130 \times 85}{100 - 85} = 736.6666666666666\) million dollars.
Step 3 ：Next, we calculate the cost of inoculating 45% of the population. Substituting \(x = 45\) into the function, we get \(C(45) = \frac{130 \times 45}{100 - 45} = 106.36363636363636\) million dollars.
Step 4 ：Finally, we subtract the cost of inoculating 45% of the population from the cost of inoculating 85% of the population to find the difference. The difference is \(736.6666666666666 - 106.36363636363636 = 630.3030303030303\) million dollars.
Step 5 ：The difference in cost between inoculating 85% of the population and inoculating 45% of the population is approximately \$630.3 million. So, the final answer is \(\boxed{630.3}\).
