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Sagot :
To solve this problem, we will be using linear equation. This is appropriate given that we only have 2 points and the problem specifies that each marble is identical in size, meaning the water rises at the same rate.
Let's identify the x and the y--our independent and dependent variables.
The independent variable here is the input--the number of marbles (5 and 12), while the dependent variable is the output--the volume reading on the beaker (22 and 50).
So we have the points (5, 22) and (12, 50).
Let's solve for the slope, m.
[tex]\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ \\ m=\frac{22-50}{5-12} \\ \\ m=\frac{-28}{-7} \\ \\ m=4 \end{gathered}[/tex]This also gives us the rise of water each time that a marble is added--4mL.
Then, we use one of the points to solve the y-intercept b. let's use (5, 22).
[tex]\begin{gathered} y=mx+b \\ \\ 22=4(5)+b \\ \\ 22=20+b \\ \\ 2=b \end{gathered}[/tex]So the equation that Martha can use to predict the height of the water is:
y = 4x + 2
Knowing this, we can solve for when x = 0 to find the height of the water before Martha added marbles.
[tex]\begin{gathered} y=4x+2 \\ y=4(0)+2 \\ y=2 \end{gathered}[/tex]The height of the water in the beaker before Martha started adding marbles was 2 mL.
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