Carter, Wyoming

From Wikipedia, the free encyclopedia

Carter is a census-designated place (CDP) in Uinta County, Wyoming, United States. The community was named after Judge William A. Carter. The population was 8 at the 2000 census.

[edit] Geography

Carter is located at 41°26′31″N, 110°25′57″W (41.442062, -110.432574)^GR1. Elevation is 6,496 feet.

According to the United States Census Bureau, the CDP has a total area of 7.9 km² (3.1 mi²), all land.

[edit] Demographics

As of the census^GR2 of 2000, there were 8 people, 4 households, and 3 families residing in the CDP. The population density was 1.0/km² (2.6/mi²). There were 6 housing units at an average density of 0.8/km² (2.0/mi²). The racial makeup of the CDP was 62.50% White, 12.50% Native American, 12.50% Asian, and 12.50% from two or more races.

There were 4 households out of which 25.0% had children under the age of 18 living with them, 75.0% were married couples living together, and 25.0% were non-families. 25.0% of all households were made up of individuals and 25.0% had someone living alone who was 65 years of age or older. The average household size was 2.00 and the average family size was 2.33.

In the CDP the population was spread out with 12.5% under the age of 18, 12.5% from 25 to 44, 62.5% from 45 to 64, and 12.5% who were 65 years of age or older. The median age was 50 years. For every 100 females there were 166.7 males. For every 100 females age 18 and over, there were 133.3 males.

The median income for a household in the CDP was $12,083, and the median income for a family was $80,488. Males had a median income of $51,250 versus $26,250 for females. The per capita income for the CDP was $27,229. None of the population or the families were below the poverty line.