Respuesta :
[tex](a)\\ \text{A charter airline finds that on its saturday flights from philadelphia to london, }\\ \text{all 120 seats will be sold if the ticket price is }\$200.\\ \\ \text{However for each }\$3\text{ increase in ticket price, the number of seats sold }\\ \text{decrease by one.}\\ \\ \text{Let the number of seats sold is s, when the ticket price is p dollars.}\\ \\ \text{at price 200 the number of seats sold is 120, so one point is }(200, 120)[/tex]
[tex]\text{and at the price of 203 the numebr of seats sold is 119, so }(203, 119)\\ \\ \text{hence the slope of linear model is}=\frac{119-120}{203-200}=-\frac{1}{3}\\ \\ \text{Hence the formula for the number of seats using point slope form is}\\ \\ s-120=-\frac{1}{3}(p-200)\\ \\ \Rightarrow s-120=-\frac{1}{3}p+\frac{200}{3}\\ \\ \Rightarrow s=-\frac{1}{3}p+\frac{200}{3}+120\\ \\ \Rightarrow s=-\frac{1}{3}p+\frac{200+360}{3}\\ \\ \Rightarrow s=-\frac{1}{3}p+\frac{560}{3}[/tex]
Hence the function is: [tex]s(p)=-\frac{1}{3}p+\frac{560}{3}[/tex]
[tex](b)\\ \text{over a certain period, the number of seats sold for this flight ranged between 70 and 115.}\\ \\ \text{so we have}\\ \\ 70\leq -\frac{1}{3}p+\frac{560}{3} \leq 115\\ \\ \Rightarrow 70\leq \frac{1}{3}(560-p) \leq 115\\ \\ \Rightarrow 210 \leq (560-p) \leq 345\\ \\ \Rightarrow 210-560 \leq -p \leq 345-560\\ \\ \Rightarrow -350 \leq -p \leq -215\\ \\ \Rightarrow 215\leq p\leq 350\\ \\ \text{hence the range of ticket prices in dollar is: }[215, \ 350][/tex]
