Seymour, Iowa

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Seymour, Iowa
City
Location of Seymour, Iowa
Location of Seymour, Iowa
Coordinates: 40°40′58″N 93°7′15″W / 40.68278°N 93.12083°W / 40.68278; -93.12083Coordinates: 40°40′58″N 93°7′15″W / 40.68278°N 93.12083°W / 40.68278; -93.12083
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 (2010)[2]
 • Total 701
 • Estimate (2016)[3] 704
 • Density 298/sq mi (115.2/km2)
Time zone Central (CST) (UTC−6)
 • Summer (DST) CDT (UTC−5)
ZIP code 52590
Area code(s) 641
FIPS code 19-71760
GNIS feature ID 0461518

Seymour is a city of Iowa in the United States. Seymour is a very tiny commune

References[change | change source]