https://file.aiquickdraw.com/mathbot/file/1725594132647poy4nzgn.jpg

To solve the problem, we first need to identify the positions of points \( C \), \( D \), and \( E \) on the number line provided.

Based on the image, the points are positioned as follows:
- \( C \) is at \(-4\)
- \( D \) is at \(-2\)
- \( E \) is at \(0\)

Now, the problem asks for the location of a point that is \(\frac{1}{3}\) of the distance from \( C \) to \( E \).

### Step-by-Step Solution:

1. **Calculate the distance from \( C \) to \( E \):**
   \[
   \text{Distance from } C \text{ to } E = E - C = 0 - (-4) = 4
   \]

2. **Find \(\frac{1}{3}\) of the distance:**
   \[
   \frac{1}{3} \times 4 = \frac{4}{3} \approx 1.33
   \]

3. **Determine the position:**
   Since we are moving from \( C \) to \( E \), starting at \(-4\) and moving to the right by \(\frac{4}{3}\):

   \[
   \text{New position} = C + \frac{4}{3} = -4 + \frac{4}{3} = -\frac{12}{3} + \frac{4}{3} = -\frac{8}{3}
   \]

So, the point is located at \(-\frac{8}{3}\), which is approximately \(-2.67\).

### Final Answer:
The location of the point that is \(\frac{1}{3}\) of the distance from \( C \) to \( E \) is \(-\frac{8}{3}\).

Would you like more details on this solution, or do you have any specific questions?

### Further Exploration Questions:
1. What would be the location if the point was \(\frac{2}{3}\) of the distance from \( C \) to \( E \)?
2. How would the answer change if point \( C \) was at \(-6\) instead of \(-4\)?
3. Can you find the point that is \(\frac{1}{4}\) of the distance from \( D \) to \( E \)?
4. What is the midpoint between points \( C \) and \( E \)?
5. How would you calculate a point that is 50% further from \( E \) than point \( C \)?

**Tip:** Always use a number line to visualize problems involving distances and fractions for more clarity.

What is the location of the point that is 1/3 of the distance from C to E on a number line?

Learn how to find the location of a point that is a fraction of the distance between two points on a number line. This problem involves calculating distances and applying fractional values. Suitable for students in Grades 6-8.

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