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The compound inequality provided is:

\[
-215 \leq x < -190
\]

This inequality can be interpreted as follows:

- \( x \) is greater than or equal to \(-215\) (closed circle on the number line at \(-215\)).
- \( x \) is strictly less than \(-190\) (open circle on the number line at \(-190\)).

### To graph the inequality on the number line:

1. **Plot a closed circle** at \(-215\) since the inequality is \( \leq \) (meaning it includes \(-215\)).
2. **Plot an open circle** at \(-190\) since the inequality is \( < \) (meaning it does not include \(-190\)).
3. **Shade the region between** \(-215\) and \(-190\) on the number line, as this is the range where \( x \) satisfies the inequality.

### Here's a step-by-step approach to graph it:

1. Place a **closed circle** at \(-215\).
2. Place an **open circle** at \(-190\).
3. Shade the segment in between, from \(-215\) to \(-190\).

Would you like further details or assistance on another part? Here are some related questions to expand understanding:

1. How do we graph inequalities that involve only one boundary, such as \(x \geq -50\)?
2. What happens when both endpoints are included, i.e., \( -10 \leq x \leq 20 \)?
3. How would you graph an inequality like \( x < -190 \)?
4. How can compound inequalities be represented in interval notation?
5. How do we solve compound inequalities that involve two variables?

**Tip**: Always check whether the inequality sign includes the boundary (\(\leq\) or \(\geq\)) to determine whether to use an open or closed circle.

Graph the compound inequality: -215 ≤ x < -190.

Learn how to graph the compound inequality -215 ≤ x < -190 on a number line. This problem involves plotting a closed circle at -215 and an open circle at -190, shading the region in between. Suitable for middle school students.

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