To obtain the amount of paper that would most likely be collected in 10 weeks, the following steps are necessary:
Step 1: Select two points that lie on the straight line and use the two points to derive the equation of the straight line, as follows:
Such two points could be: (x1, y1) = (0, 30) and (x2, y2) = (120, 3)
Using the following formula, we can derive the equation of the straight line:
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]
Thus:
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \Rightarrow\frac{y-30_{}}{x-0_{}}=\frac{120_{}-30_{}}{3_{}-0_{}} \\ \Rightarrow\frac{y-30_{}}{x_{}}=\frac{90_{}}{3_{}_{}} \\ \Rightarrow\frac{y-30_{}}{x_{}}=30 \\ \Rightarrow y-30=30\times x \\ \Rightarrow y=30x+30 \end{gathered}[/tex]
The above equation can be re-written as:
[tex]\begin{gathered} y=30x+30 \\ \Rightarrow\text{Amount of paper collected = 30 }\times number\text{ of w}eeks\text{ + 30 } \end{gathered}[/tex]
Step 2: Use the derived equation to obtain the value of the amount of paper collected in 10 weeks, as follows:
In 10 weeks, we will have :
[tex]\begin{gathered} \text{Amount of paper collected = 30 }\times number\text{ of w}eeks\text{ + 30 } \\ \Rightarrow\text{Amount of paper collected = 30 }\times10\text{ + 30 } \\ \Rightarrow\text{Amount of paper collected = 300 + 30 }=330 \\ \Rightarrow\text{Amount of paper collected = 3}30 \end{gathered}[/tex]
Therefore, the amount of paper that would most likely be collected in 10 weeks is 330
