Jefferson, Ohio

From Simple English Wikipedia, the free encyclopedia
Jump to navigation Jump to search
Jefferson, Ohio
Ashtabula County Courthouse
Location of Jefferson, Ohio
Location of Jefferson, Ohio
Location of Jefferson in Ashtabula County
Location of Jefferson in Ashtabula County
Coordinates: 41°44′14″N 80°46′19″W / 41.73722°N 80.77194°W / 41.73722; -80.77194Coordinates: 41°44′14″N 80°46′19″W / 41.73722°N 80.77194°W / 41.73722; -80.77194
CountryUnited States
StateOhio
CountyAshtabula
Government
 • MayorJudy Maloney
Area
 • Total2.52 sq mi (6.53 km2)
 • Land2.52 sq mi (6.53 km2)
 • Water0 sq mi (0 km2)
Elevation958 ft (292 m)
Population
 (2010)
 • Total3,120
 • Estimate 
(2018[2])
2,994
 • Density1,238.1/sq mi (478.0/km2)
Time zoneUTC-5 (Eastern (EST))
 • Summer (DST)UTC-4 (EDT)
ZIP code
44047
Area code(s)440
FIPS code39-38500[3]
GNIS feature ID1042061
Websitehttp://www.jeffersonohio.us/

Jefferson is a village in Ashtabula County, Ohio, United States. The population was 3,120 at the 2010 census. It is the county seat of Ashtabula County.[4]

References[change | change source]

  1. "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
  2. "Population and Housing Unit Estimates". Retrieved June 30, 2019.
  3. "U.S. Census website". United States Census Bureau. Retrieved 2008-01-31.
  4. "Find a County". National Association of Counties. Archived from the original on 2011-05-31. Retrieved 2011-06-07.