Seymour, Iowa
Seymour is a city in Iowa in the United States. Seymour is a very tiny city.
City | |
 Location of Seymour, Iowa | |
| Coordinates: 40°40′58″N 93°7′15″W / 40.68278°N 93.12083°WCoordinates: 40°40′58″N 93°7′15″W / 40.68278°N 93.12083°W | |
| Country | USA |
| State |  Iowa |
| County | Wayne |
| Area .[1] | |
| • Total | 2.35 sq mi (6.09 km2) |
| • Land | 2.35 sq mi (6.09 km2) |
| • Water | 0 sq mi (0 km2) |
| Elevation | 1,060 ft (323 m) |
| Population | |
| • Total | 701 |
| • Estimate (2016)[3] | 704 |
| • Density | 298/sq mi (115.2/km2) |
| Time zone | UTC−6 (Central (CST)) |
| • Summer (DST) | UTC−5 (CDT) |
| ZIP code | 52590 |
| FIPS code | 19-71760 |
| GNIS feature ID | 0461518 |
References
- ↑ "US Gazetteer files 2010". United States Census Bureau. Retrieved 2012-05-11.
- ↑ "American FactFinder". United States Census Bureau. Retrieved 2012-05-11.
- ↑ "Population and Housing Unit Estimates". Retrieved June 9, 2017.
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