Reid Senesac M

University Of Iowa
Nursing Assistant

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Reid Senesac M Overview

Reid Senesac M in 2018 was employed at University Of Iowa and had a reported pay of $25,176, according to public records. This pay was 60 percent lower than the average and 52 percent lower than the median salary in University Of Iowa.

University Of Iowa records show Reid Senesac M held two jobs from 2018 to 2019.

In year 2018 Reid Senesac's salary was 450 percent higher than average Nursing Assistant salary in the state of Iowa.

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Key Data

Year2018
Full NameReid Senesac M
Job TitleNursing Assistant
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StateIowa
EmployerUniversity Of Iowa
Total Pay $25,176

*Information may include where available: salary, bonuses, benefits, retirement contributions, pensions, and other financial data.

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Employer

Employer NameUniversity Of Iowa
Year2018
Number of Employees24,328
Average Annual Wage$63,150
Median Annual Wage$51,979
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Other records for Reid Senesac M

 (2019)
Research Associate
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Highest Paid with the Same Job Title (Nursing Assistant)

Co-Workers

 (2018)
Visiting Associate at University Of Iowa
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 (2018)
Clerk at University Of Iowa
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 (2018)
Clerk at University Of Iowa
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 (2018)
Clinical Professor at University Of Iowa
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 (2018)
Cook at University Of Iowa
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 (2018)
Psychiatric Nursing Asst at University Of Iowa
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 (2018)
Custodian at University Of Iowa
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 (2018)
Adjunct Associate Professor at University Of Iowa
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 (2018)
Clerk at University Of Iowa
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 (2018)
Administrative Services Coordinator at University Of Iowa
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