Question

this situation can be modeled as a linear relationship.
complete the statement that describes the situation.
mrs. kerr prepares pounds of clay and hands out

Explanation:

Step1: Find initial clay (y-intercept)

The y - intercept is when \( x = 0 \) (number of students is 0). From the graph, when \( x = 0 \), \( y = 45 \). So Mrs. Kerr prepares 45 pounds of clay.

Step2: Calculate clay per student (slope)

Slope \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Take two points: \( (0, 45) \) and \( (30, 5) \).
\( m=\frac{5 - 45}{30 - 0}=\frac{- 40}{30}=-\frac{4}{3}\)? Wait, no, wait. Wait, when \( x = 30 \), \( y = 5 \). Wait, let's check another point. Wait, when \( x = 0 \), \( y = 45 \); when \( x = 30 \), \( y = 5 \). The change in \( y \) is \( 5 - 45=-40 \), change in \( x \) is \( 30 - 0 = 30 \). But maybe a better way: the slope represents the amount of clay handed out per student (negative because it's decreasing). Wait, actually, the slope \( m=\frac{\Delta y}{\Delta x}=\frac{y_2 - y_1}{x_2 - x_1} \). Let's take \( (0,45) \) and \( (30,5) \). So \( m=\frac{5 - 45}{30 - 0}=\frac{-40}{30}=-\frac{4}{3}\)? Wait, that can't be. Wait, maybe I misread the graph. Wait, the y - axis is clay remaining. So when there are 0 students, clay remaining is 45 (so initial clay is 45). When there are 30 students, clay remaining is 5. So the total clay handed out in 30 students is \( 45 - 5 = 40 \) pounds. So per student, it's \( \frac{40}{30}=\frac{4}{3}\)? Wait, no, the slope is \( \frac{\text{change in }y}{\text{change in }x}=\frac{5 - 45}{30 - 0}=\frac{-40}{30}=-\frac{4}{3} \). The negative sign means that for each student, the clay remaining decreases by \( \frac{4}{3} \) pounds, which means she hands out \( \frac{4}{3} \) pounds per student? Wait, no, wait, maybe I made a mistake. Wait, let's check the graph again. Wait, the y - axis is clay remaining (pounds), x - axis is number of students. So the equation of the line is \( y=mx + b \), where \( b = 45 \) (y - intercept). Let's take another point. When \( x = 15 \), what's \( y \)? From the graph, when \( x = 15 \), \( y = 25 \)? Wait, no, the grid: each square is, let's see, x - axis: 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. y - axis: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. Wait, when \( x = 0 \), \( y = 45 \); when \( x = 30 \), \( y = 5 \). So the slope \( m=\frac{5 - 45}{30 - 0}=\frac{-40}{30}=-\frac{4}{3}\). So the equation is \( y=-\frac{4}{3}x + 45 \). So the amount of clay handed out per student is the absolute value of the slope, which is \( \frac{4}{3} \) pounds per student? Wait, but maybe the problem is simpler. The initial clay is 45 pounds (when \( x = 0 \), \( y = 45 \)). The total clay handed out when \( x = 30 \) is \( 45 - 5 = 40 \) pounds, so per student, \( \frac{40}{30}=\frac{4}{3}\approx1.33 \) pounds? Wait, but maybe the graph is such that when \( x = 0 \), \( y = 45 \) (initial clay), and the slope is \( - \frac{4}{3} \), meaning that for each student, \( \frac{4}{3} \) pounds of clay are handed out (so clay remaining decreases by \( \frac{4}{3} \) per student). But maybe the problem just wants the initial clay (45 pounds) and the rate (slope's absolute value or the amount per student). Wait, the problem says "hands out" how many pounds per student. So from \( x = 0 \) to \( x = 30 \), clay remaining goes from 45 to 5, so total handed out is \( 45 - 5 = 40 \) pounds over 30 students. So per student, \( \frac{40}{30}=\frac{4}{3}\) pounds? Wait, no, that seems odd. Wait, maybe I misread the graph. Wait, maybe when \( x = 30 \), \( y = 0 \)? No, the graph shows \( y = 5 \) at \( x = 30 \). Wait, maybe the y - axis is from 0 to 50, with 5 - unit intervals. Wait, the first point is (0,45), then when x = 3, y = 40? Wait, let's recalculate the slope. Let's take (0,45) and (3,40). Then \( m=\frac{40 - 45}{3 - 0}=\frac{-5}{3}\approx - 1.666 \). Then (3,40) and (6,35): \( m=\frac{35 - 40}{6 - 3}=\frac{-5}{3}\). Ah! So the slope is \( -\frac{5}{3} \). Wait, I see, I misread the graph earlier. The y - axis: 45, 40, 35, 30, 25, 20, 15, 10, 5. So each grid line on y is 5 units. So when x = 0, y = 45; x = 3, y = 40; x = 6, y = 35; x = 9, y = 30; x = 12, y = 25; x = 15, y = 20; x = 18, y = 15; x = 21, y = 10; x = 24, y = 5; x = 27, y = 0? Wait, no, the last point is at x = 30, y = 5? No, maybe the last point is at x = 24, y = 5? Wait, the graph is a straight line. Let's take two clear points: (0,45) and (24,5). Then slope \( m=\frac{5 - 45}{24 - 0}=\frac{-40}{24}=-\frac{5}{3}\). Ah, that makes sense. So from x = 0 to x = 24, y goes from 45 to 5, change in y is - 40, change in x is 24. So \( m =-\frac{40}{24}=-\frac{5}{3}\). So the equation is \( y =-\frac{5}{3}x + 45 \). So when x = 24, \( y=-\frac{5}{3}(24)+45=-40 + 45 = 5 \), which matches. So the slope is \( -\frac{5}{3} \), meaning that for each student, the clay remaining decreases by \( \frac{5}{3} \) pounds, so she hands out \( \frac{5}{3} \) pounds per student? Wait, no, the slope is \( \frac{\text{change in }y}{\text{change in }x} \). Since y is clay remaining, a negative slope means that as x (students) increases, y (clay remaining) decreases. So the rate of change of clay remaining is \( -\frac{5}{3} \) pounds per student, which means the rate of clay handed out is \( \frac{5}{3} \) pounds per student. But the initial amount of clay is 45 pounds (when x = 0, y = 45).

Answer:

Mrs. Kerr prepares 45 pounds of clay and hands out \(\frac{5}{3}\) (or approximately 1.67) pounds of clay per student. (If we consider the slope calculation with points (0,45) and (24,5), the slope is \(-\frac{5}{3}\), so the amount handed out per student is \(\frac{5}{3}\) pounds.)