is a town located in Sweetwater County, Wyoming
. As of the 2000
census, the town had a total population of 17.
Geography\nSweeney Ranch is located at 41°28'23" North, 109°0'52" West (41.473176, -109.014468)1
According to the United States Census Bureau
, the town has a total area of 21.0 km²
). 21.0 km² (8.1 mi²) of it is land and none of the area is covered with water.
Demographics\nAs of the census
, there are 17 people, 6 households, and 4 families residing in the town. The population density
is 0.8/km² (2.1/mi²). There are 7 housing units at an average density of 0.3/km² (0.9/mi²). The racial makeup of the town is 100.00% White, 0.00% African American
, 0.00% Native American
, 0.00% Asian
, 0.00% Pacific Islander, 0.00% from other races, and 0.00% from two or more races. 0.00% of the population are Hispanic
of any race.
There are 6 households out of which 33.3% have children under the age of 18 living with them, 66.7% are married couples living together, 16.7% have a female householder with no husband present, and 16.7% are non-families. 16.7% of all households are made up of individuals and 0.0% have someone living alone who is 65 years of age or older. The average household size is 2.83 and the average family size is 3.20.
In the town the population is spread out with 17.6% under the age of 18, 17.6% from 18 to 24, 11.8% from 25 to 44, 52.9% from 45 to 64, and 0.0% who are 65 years of age or older. The median age is 46 years. For every 100 females there are 112.5 males. For every 100 females age 18 and over, there are 100.0 males.
The median income for a household in the town is $31,250, and the median income for a family is $0. Males have a median income of $31,250 versus $0 for females. The per capita income for the town is $31,000. 0.0% of the population and 0.0% of families are below the poverty line.