Gerald Layton
Instructional Lab Technician (2019)
| Regular pay: | $9,709.86 |
|---|---|
| Overtime pay: | $0.00 |
| Other pay: | $0.00 |
| Total pay: | $9,709.86 |
| Benefits: | $0.00 |
| Total pay & benefits: | $9,709.86 |
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