Warrensville Heights, Ohio

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Warrensville Heights is a city in Cuyahoga County, Ohio, United States. The population was 15,109 at the 2000 census.



Warrensville Heights is located at 41°26′19″N 81°31′24″W / 41.43861°N 81.52333°W / 41.43861; -81.52333 (41.438653, -81.523262)[4].

According to the United States Census Bureau, the city has a total area of 4.1 square miles (10.7 km²), all of it land.


As of the census[2] of 2000, there were 15,109 people, 6,325 households, and 4,048 families residing in the city. The population density was 3,661.4 people per square mile (1,412.5/km²). There were 6,741 housing units at an average density of 1,633.5/sq mi (630.2/km²). The racial makeup of the city was 6.57% White, 90.41% African American, 0.17% Native American, 0.91% Asian, 0.04% Pacific Islander, 0.32% from other races, and 1.58% from two or more races. Hispanic or Latino of any race were 0.75% of the population.

There were 6,325 households out of which 28.6% had children under the age of 18 living with them, 32.2% were married couples living together, 27.7% had a female householder with no husband present, and 36.0% were non-families. 32.4% of all households were made up of individuals and 9.2% had someone living alone who was 65 years of age or older. The average household size was 2.34 and the average family size was 2.95.

In the city the population was spread out with 25.6% under the age of 18, 7.9% from 18 to 24, 26.6% from 25 to 44, 25.5% from 45 to 64, and 14.4% who were 65 years of age or older. The median age was 38 years. For every 100 females there were 73.7 males. For every 100 females age 18 and over, there were 66.8 males.

The median income for a household in the city was $37,204, and the median income for a family was $41,962. Males had a median income of $35,947 versus $28,422 for females. The per capita income for the city was $18,611. About 10.4% of families and 11.4% of the population were below the poverty line, including 16.2% of those under age 18 and 9.5% of those age 65 or over.

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