Solution:
Given the scatterplot below:
where the line of best fit is expressed as
![y=0.91x+7.99](https://img.qammunity.org/2023/formulas/mathematics/college/we9negoweetojmha2mr0kwtoeoaq3ev8ks.png)
A) Predicted hourly rate for cashier with 9 years of experience:
Thus, we have
![x=9](https://img.qammunity.org/2023/formulas/mathematics/high-school/cmoob0b43q6h3m27uzhcp4tu0db7ka7bi8.png)
By substituting the value of 9 for x into the equation, we have
![\begin{gathered} y=0.91\left(9\right)+7.99 \\ =\$16.18 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cjd8q2gknhgamw9eeuurvlrvnhx8h2q08d.png)
B) Predicted hourly rate for cashier with no experience:
This implies that
![x=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/fbdfo5wsa562jve9mar3rnrxezpj37nli7.png)
By substitution, we have
![\begin{gathered} y=0.91(0)+7.99 \\ \Rightarrow y=\$7.99 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x5n7n4kmoythcy2u1ig3b2f36bu8o11rqc.png)
C) Predicted increase in the hourly rate for an increase of one year of experience:
Recall that the equation of a line is expressed as
![\begin{gathered} y=mx+c \\ where \\ m\Rightarrow slope \\ m=\frac{increase\text{ in y}}{increase\text{ in x}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/udybsjjphntkj2luaa65yvhgef3l0g2tgr.png)
From the equation of the line of best fit, by comparison, we have
![\begin{gathered} slope=0.91 \\ where \\ slope=\frac{incresre\text{ in hourly pay}}{increase\text{ in year of experince}} \\ \Rightarrow0.91=\frac{increase\text{ in hourly pay}}{1} \\ thus,\text{ we have} \\ predicted\text{ increase in hourly pay = \$0.91} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rowyesgdaszl69ajnvdi582nk7w38jc66g.png)