Cuba, New York
Cuba is a town in Allegany County, New York, United States. As of the 2020 United States Census, the total population was 3,126.[2]
 | |
| Coordinates: 42°13′04″N 78°16′31″W / 42.21778°N 78.27528°WCoordinates: 42°13′04″N 78°16′31″W / 42.21778°N 78.27528°W | |
| Country | United States |
| State | New York |
| County | Allegany |
| Area | |
| • Total | 35.80 sq mi (92.73 km2) |
| • Land | 35.10 sq mi (90.91 km2) |
| • Water | 0.70 sq mi (1.82 km2) |
| Population | |
| • Total | 3,126 |
| • Density | 89.06/sq mi (34.39/km2) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| FIPS code | 36-003-19367 |
| Website | https://www.cubany.org/ |
References
- ↑ "2020 U.S. Gazetteer Files". United States Census Bureau. Retrieved March 7, 2024.
- ↑ 2.0 2.1 "Profile of General Population and Housing Characteristics: 2020 Demographic Profile Data (DP-1): Cuba town, Allegany County, New York". United States Census Bureau. Retrieved March 7, 2024.
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