To determine whether any of the given operations result in an arithmetic overflow for the 8-bit signed integers represented in sign-magnitude form, let's analyze each operation.

### Understanding Sign-Magnitude Representation:

• An 8-bit sign-magnitude number uses the first bit as the sign bit (0 for positive, 1 for negative) and the remaining 7 bits for magnitude.
• The range for 8-bit signed integers is -127 to +127.

### Given:

• \(X = 10110100\). The sign bit is 1 (negative), and the magnitude is \(0110100 = 52\).
• Thus, \(X = -52\).
• \(Y = 01001100\). The sign bit is 0 (positive), and the magnitude is \(1001100 = 76\).
• Thus, \(Y = +76\).

### Analyzing Each Operation:

1. \(Z = X + Y\)
   • Value: \(Z = -52 + 76 = 24\)
   • 24 is within the valid range, hence no overflow occurs.
2. \(Z = X - Y\)
   • Value: \(Z = -52 - 76 = -128\)
   • -128 is outside the valid range for 8-bit signed numbers, hence overflow occurs.
3. \(Z = -X + Y\)
   • Negation of \(X\) (since \(X = -52\), thus \(-X = 52\))
   • Value: \(Z = 52 + 76 = 128\)
   • 128 is outside the valid range, hence overflow occurs.
4. \(Z = -X - Y\)
   • Negation of \(X\) gives \(52\).
   • Value: \(Z = 52 - 76 = -24\)
   • -24 is within the valid range, hence no overflow occurs.

### Conclusion:

• Operations that result in overflow are \(Z = X - Y\) and \(Z = -X + Y\).