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To solve the problem of finding the constant of proportionality between the number of hours Dale spends babysitting (denoted as $x$) and the amount of money he earns (denoted as $y$), we use the formula:

\[
k = \frac{y}{x}
\]

This formula defines the constant of proportionality $k$, where $k$ is the amount Dale earns per hour.

From the table provided:

\[
\begin{array}{|c|c|}
\hline
x \ (\text{hours}) & y \ (\text{dollars}) \\
\hline
1 & 5 \\
2 & 10 \\
3 & 15 \\
4 & 20 \\
\hline
\end{array}
\]

Now, calculate the constant of proportionality $k$ for each data point:

- For $x = 1$ and $y = 5$:
  \[
  k = \frac{5}{1} = 5
  \]

- For $x = 2$ and $y = 10$:
  \[
  k = \frac{10}{2} = 5
  \]

- For $x = 3$ and $y = 15$:
  \[
  k = \frac{15}{3} = 5
  \]

- For $x = 4$ and $y = 20$:
  \[
  k = \frac{20}{4} = 5
  \]

The constant of proportionality is consistent for all values of $x$ and $y$, and it is $k = 5$. This means Dale earns 5 dollars per hour.

Would you like any further clarification?

---

Here are 5 related questions you could explore:
1. What would Dale earn if he worked for 6 hours?
2. How many hours does Dale need to babysit to earn $40?
3. Can you write the equation that models the relationship between $x$ (hours) and $y$ (dollars)?
4. If Dale wants to save $100, how many hours must he babysit?
5. Is this relationship linear, and how can you tell?

**Tip**: In any proportional relationship, dividing the two related quantities (y/x) always gives you the constant of proportionality.

Dale wants to save up some money to buy a new smartphone, so he babysits on the weekends. There is a proportional relationship between the time Dale spends babysitting (in hours), x, and the amount of money he earns babysitting (in dollars), y. What is the constant of proportionality? Write your answer as a whole number or decimal.

Learn how to determine the constant of proportionality in a babysitting earnings problem. Explore the relationship between time spent babysitting and earnings using a linear model. Suitable for Grades 6-8.

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