Question

part 3 of 3 after filling the ketchup dispenser at the snack bar where she works, kelley measures the level of ketchup during the day at different hourly intervals. complete parts a to c. b. how can the equation from part a be used to find the level of ketchup when the dispenser is full? a. substitute 0 for y in the equation and solve for x. equivalently, find the x-intercept. b. substitute 0 for x in the equation and solve for y. equivalently, find the y-intercept. c. identify the slope of the line represented by the equation. d. find the difference between the slope and the y-intercept of the line represented by the equation. c. if kelley fills the ketchup dispenser just before the snack bar opens and the snack bar is open for 18 hours, will the dispenser need to be refilled before closing time? explain. if the ketchup is used at a constant rate, then the equation from part a indicates the dispenser will become empty \boxed{} hours after the snack bar opens. this means that the dispenser \boxed{} and so it \boxed{} need to be refilled before closing time.

Explanation:

Part b

To find the level of ketchup when the dispenser is full, we consider the time when the snack bar just opens (x = 0, where x is time in hours). The equation from part a (presumably a linear equation \(y = mx + b\), where \(y\) is ketchup level and \(x\) is time) will have \(b\) as the initial level (full level) when \(x = 0\). So substituting \(x = 0\) and solving for \(y\) gives the y - intercept, which is the full level. Option A is for when \(y = 0\) (empty), option C is about slope (rate of change, not full level), option D is incorrect. So option B is correct.

Step 1: Find the slope \(m\)

The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Using points \((5,\frac{95}{8})\) and \((8,10)\) (where \(10=\frac{80}{8}\)):
\(m=\frac{\frac{80}{8}-\frac{95}{8}}{8 - 5}=\frac{-\frac{15}{8}}{3}=-\frac{15}{8}\times\frac{1}{3}=-\frac{5}{8}\)

Step 2: Find the y - intercept \(b\)

Using the point \((8,10)\) and \(m =-\frac{5}{8}\) in \(y=mx + b\):
\(10=-\frac{5}{8}\times8 + b\)
\(10=- 5 + b\)
\(b = 15\)

So the equation is \(y=-\frac{5}{8}x + 15\)

Step 3: Find when \(y = 0\) (dispenser empty)

Set \(y = 0\) in \(y=-\frac{5}{8}x + 15\):
\(0=-\frac{5}{8}x+15\)
\(\frac{5}{8}x = 15\)
\(x=15\times\frac{8}{5}=24\)

The snack bar is open for 18 hours. Since \(24>18\), the dispenser becomes empty 24 hours after opening, and since 24 hours is more than 18 hours, the dispenser will be empty after closing time, so it does not need to be refilled before closing time.

Answer:

B. Substitute 0 for x in the equation and solve for y. Equivalently, find the y - intercept.

Part c

First, we need to find the equation of the line from the given data points. Let's assume the linear equation is \(y=mx + b\), where \(x\) is time in hours and \(y\) is ketchup level in inches.

We have three points: \((5,11\frac{7}{8})=(5,\frac{95}{8})\), \((8,10)\), \((11,8\frac{1}{8})=(11,\frac{65}{8})\)