(a)
Let Wj be the wages of John and Tj be the time in hours.
If wages and time are proportional, then the relation between them can be expressed as,
![W_j=kT_j](https://img.qammunity.org/2023/formulas/mathematics/college/c6hxjaltns7gvtlpty1x5sg9vzifxkl3m8.png)
Here, k is a constant
Rewriting the above equation,
![k=(W_j)/(T_j)](https://img.qammunity.org/2023/formulas/mathematics/college/xqctszm1xf2e7ihv6lcq0ww48orixh7cao.png)
If the ratio of wages to hours is a constant, then the wages of John is proportional to time.
Taking the ratio of Wj to Tj, we get
![\begin{gathered} k_{}=(18)/(2)=9 \\ _{}k=(27)/(3)=9 \\ k=(36)/(4)=9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/unj972bo2h8p3kppkl1ch8hufg9wc5dqku.png)
Since the ratio of wages to time is a constant, John's wages are proportional to time.
The constant k is the the unit rate. Therefore, the unit rate at which John works is $9/h.
(b)
The graph plotted with Amber's wages against hours worked is a straight line. Therefore, Amber's wages are proportional to time.
The slope of the graph gives the unit rate.
Hence, the unit rate at which Amber work is given by,
![\begin{gathered} \text{Unit Rate=Slope} \\ =(\Delta y)/(\Delta x) \\ =(16-0)/(2-0) \\ =\text{ \$8/h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ug5szapaxgplxsvy2ay19c9sw5lojjs12u.png)
Therefore, the unit rate at which Amber work is $8/h