Rushville, Nebraska

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Rushville is a city in Sheridan County, Nebraska, United States. The population was 999 at the 2000 census. It is the county seat of Sheridan County[3].



Rushville is located at 42°43′2″N 102°27′37″W / 42.71722°N 102.46028°W / 42.71722; -102.46028 (42.717209, -102.460350)[4].

According to the United States Census Bureau, the city has a total area of 1.2 square miles (3.0 km²), all of it land.


As of the census[1] of 2000, there were 999 people, 419 households, and 266 families residing in the city. The population density was 857.2 people per square mile (329.7/km²). There were 518 housing units at an average density of 444.5/sq mi (170.9/km²). The racial makeup of the city was 77.68% White, 0.10% African American, 18.32% Native American, 0.30% from other races, and 3.60% from two or more races. Hispanic or Latino of any race were 2.30% of the population.

There were 419 households out of which 28.9% had children under the age of 18 living with them, 47.5% were married couples living together, 11.7% had a female householder with no husband present, and 36.3% were non-families. 33.7% of all households were made up of individuals and 16.5% had someone living alone who was 65 years of age or older. The average household size was 2.30 and the average family size was 2.90.

In the city the population was spread out with 25.5% under the age of 18, 5.9% from 18 to 24, 25.3% from 25 to 44, 22.3% from 45 to 64, and 20.9% who were 65 years of age or older. The median age was 41 years. For every 100 females there were 88.1 males. For every 100 females age 18 and over, there were 86.5 males.

The median income for a household in the city was $27,361, and the median income for a family was $36,100. Males had a median income of $23,750 versus $20,536 for females. The per capita income for the city was $14,586. About 13.5% of families and 16.4% of the population were below the poverty line, including 28.7% of those under age 18 and 7.0% of those age 65 or over.

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