Question 18 At $$ \$ 350 $$ per person, an airline anticipates selling 250 tickets for a particular flight. At $$ \$ 450 $$ per person, the airline anticipates selling 100 tickets for the same flight. Assume a linear relation between the cost per ticket $$ C $$ and the number of tickets, $$ x $$ sold. Which of the following equations can be used to model the given information? OC=-1 $$ x+521.66666666667 $$ OC=-1 $$ x+516.66666666667 $$ OC=- $$ \frac{2}{3} x+521.66666666666 $$ Question Help: Video D Post to forum Submit Question

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UpStudy AI · Accepted Answer

We begin by letting $$ x $$ be the number of tickets sold and $$ C $$ the cost per ticket. We are given two points on the line:

$$
(250,\,350) \quad \text{and} \quad (100,\,450)
$$

Step 1. Compute the slope $$ m $$.

$$
m=\frac{450-350}{100-250}=\frac{100}{-150}=-\frac{2}{3}
$$

Step 2. Write the equation in point–slope form using the point $$ (250,\,350) $$:

$$
C-350=-\frac{2}{3}\,(x-250)
$$

Step 3. Solve for $$ C $$:

$$
\begin{aligned}
C &= -\frac{2}{3}\,x+\frac{2}{3}\cdot250+350\[1mm]
  &= -\frac{2}{3}\,x+\frac{500}{3}+350
\end{aligned}
$$

Step 4. Express 350 as a fraction with denominator 3:

$$
350=\frac{1050}{3}
$$

Now combine the terms:

$$
C=-\frac{2}{3}\,x+\frac{500+1050}{3}=-\frac{2}{3}\,x+\frac{1550}{3}
$$

Since

$$
\frac{1550}{3}\approx516.67,
$$

the model is

$$
C=-\frac{2}{3}\,x+516.667.
$$

Of the answer choices given, the only option with the correct slope is

$$
\boxed{C=-\frac{2}{3}\,x+521.66666666666},
$$

which differs slightly in the intercept (521.67 versus 516.67). Since the slope must be $$-\frac{2}{3}$$ to match the data and none of the choices exactly equal $$-\frac{2}{3}\,x+516.667$$, the best answer among the choices is the one with slope $$-\frac{2}{3}$$:

$$
\boxed{C=-\frac{2}{3}\,x+521.66666666666.}
$$
The correct equation to model the given information is: $$ C = -\frac{2}{3}x + 521.66666666666 $$

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