Converse, Texas

From Wikipedia, the free encyclopedia

Converse is a city in Bexar County, Texas, United States. The population was 11,508 at the 2000 census.

[edit] Geography

Converse is located at 29°31′2″N, 98°18′50″W (29.517280, -98.313804)^GR1.

According to the United States Census Bureau, the city has a total area of 16.4 km² (6.3 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 11,508 people, 3,837 households, and 3,077 families residing in the city. The population density was 701.9/km² (1,818.3/mi²). There were 4,009 housing units at an average density of 244.5/km² (633.4/mi²). The racial makeup of the city was 69.39% White, 13.03% African American, 0.56% Native American, 2.20% Asian, 0.14% Pacific Islander, 10.65% from other races, and 4.04% from two or more races. Hispanic or Latino of any race were 29.44% of the population.

There were 3,837 households out of which 44.6% had children under the age of 18 living with them, 61.5% were married couples living together, 14.5% had a female householder with no husband present, and 19.8% were non-families. 15.5% of all households were made up of individuals and 3.2% had someone living alone who was 65 years of age or older. The average household size was 2.94 and the average family size was 3.26.

In the city the population was spread out with 31.0% under the age of 18, 8.0% from 18 to 24, 33.9% from 25 to 44, 20.8% from 45 to 64, and 6.4% who were 65 years of age or older. The median age was 32 years. For every 100 females there were 95.5 males. For every 100 females age 18 and over, there were 89.2 males.

The median income for a household in the city was $47,947, and the median income for a family was $49,396. Males had a median income of $32,631 versus $25,765 for females. The per capita income for the city was $18,949. About 4.9% of families and 6.5% of the population were below the poverty line, including 8.5% of those under age 18 and 10.5% of those age 65 or over.