Henry, South Dakota

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Henry is a town in Codington County, South Dakota, United States. The population was 268 at the 2000 census.

[edit] Geography

Henry is located at 44°52′40″N, 97°27′57″W (44.877650, -97.465968)^GR1.

According to the United States Census Bureau, the town has a total area of 3.8 km² (1.5 mi²), all land.

Henry has been assigned the ZIP code 57243 and the FIPS place code 28140.

[edit] Demographics

As of the census^GR2 of 2000, there were 268 people, 101 households, and 71 families residing in the town. The population density was 71.4/km² (185.3/mi²). There were 116 housing units at an average density of 30.9/km² (80.2/mi²). The racial makeup of the town was 98.88% White, 0.37% Native American, 0.75% from other races. Hispanic or Latino of any race were 1.87% of the population.

There were 101 households out of which 37.6% had children under the age of 18 living with them, 57.4% were married couples living together, 9.9% had a female householder with no husband present, and 29.7% were non-families. 26.7% of all households were made up of individuals and 9.9% had someone living alone who was 65 years of age or older. The average household size was 2.65 and the average family size was 3.28.

In the town the population was spread out with 28.7% under the age of 18, 8.6% from 18 to 24, 31.7% from 25 to 44, 21.3% from 45 to 64, and 9.7% who were 65 years of age or older. The median age was 32 years. For every 100 females there were 111.0 males. For every 100 females age 18 and over, there were 103.2 males.

The median income for a household in the town was $36,607, and the median income for a family was $42,500. Males had a median income of $24,167 versus $25,750 for females. The per capita income for the town was $13,778. About 5.1% of families and 10.0% of the population were below the poverty line, including 10.0% of those under the age of eighteen and 38.5% of those sixty five or over.