Which set expresses the solution to the compound inequality −2≤2x+13−4<1−2≤2x+13−4<1?(1 point)

To solve the compound inequality \(-2 \leq 2x + 13 - 4 < 1\), we can break it into two parts:

1. \(-2 \leq 2x + 13 - 4\)
2. \(2x + 13 - 4 < 1\)

First, simplify the expression:

\[
2x + 13 - 4 = 2x + 9
\]

Now we have the compound inequality:

\[
-2 \leq 2x + 9 < 1
\]

Now, let's solve both parts.

### Solve the left part:
\[
-2 \leq 2x + 9
\]
Subtract 9 from both sides:

\[
-2 - 9 \leq 2x \implies -11 \leq 2x
\]
Divide by 2:

\[
-\frac{11}{2} \leq x \implies x \geq -\frac{11}{2}
\]

### Solve the right part:
\[
2x + 9 < 1
\]
Subtract 9 from both sides:

\[
2x < 1 - 9 \implies 2x < -8
\]
Divide by 2:

\[
x < -4
\]

### Combine the results:
So now we have:

\[
-\frac{11}{2} \leq x < -4
\]

### Convert \(-\frac{11}{2}\) to decimal:
\(-\frac{11}{2} = -5.5\).

Thus, the solution can be written in interval notation as:

\[
[-\frac{11}{2}, -4)
\]

Now let's check the given options. You specifically asked for an option that matches:

- \(\{x: -\frac{11}{2} \leq x < -4\}\)

### Which response matches?
None of the given responses match our derived solution.

The correct answer would be the set:

\(\{x: -\frac{11}{2} \leq x < -4 \}\).

If you meant to have an option that I didn't see or if there was any confusion, please clarify, and I can help you further!