Solve the problem.

- vendor has learned that, by pricing pretzels at $1.50, sales will reach 109 pretzels per day. Raising the price to $2.50 will cause the sales to fall to 69 pretzels per day. Let y be the number of pretzels the vendor sells at x dollars each. Write a linear equation that models the number of pretzels sold per day (y) when the price is x dollars each.
A) $ y=40 x+49 $
B) $ y=-\frac{1}{40} x+\frac{8717}{80} $
C) $ y=-40 x-169 $
D) $ y=-40 x+169 $

Question

Solve the problem.

- vendor has learned that, by pricing pretzels at $1.50, sales will reach 109 pretzels per day. Raising the price to $2.50 will cause the sales to fall to 69 pretzels per day. Let y be the number of pretzels the vendor sells at x dollars each. Write a linear equation that models the number of pretzels sold per day (y) when the price is x dollars each.
A) $ y=40 x+49 $
B) $ y=-\frac{1}{40} x+\frac{8717}{80} $
C) $ y=-40 x-169 $
D) $ y=-40 x+169 $

Quizplus · Accepted Answer

We know that when the price is $1.50, the vendor sells 109 pretzels per day, and when the price is $2.50, the vendor sells 69 pretzels per day.

We can use this information to find the equation of a line that passes through these two points.

First, let's find the slope of the line:

slope = (change in y)/(change in x)

slope = (69-109)/($2.50-$1.50)

slope = -40

Now we can use the point-slope form of a linear equation:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is one of the points on the line.

Let's use the point (109, $1.50):

y - 109 = -40(x - 1.5)

y - 109 = -40x + 60

y = -40x + 169

So the linear equation that models the number of pretzels sold per day is y = -40x + 169.

Therefore, the answer is D.

Solve the Problem y=40x+49 y=40 x+49 y=40x+49 B) y=−140x+871780 y=-\frac{1}{40} x+\frac{8717}{80} y=−401​x+808717​

Solve the Problem $y=40 x+49$ B) $y=-\frac{1}{40} x+\frac{8717}{80}$