Poplar, Montana

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Poplar is a town in Roosevelt County, Montana, United States. The population was 911 at the 2000 census.

Contents

Geography

Poplar is located at 48°6′34″N 105°11′42″W / 48.10944°N 105.195°W / 48.10944; -105.195 (48.109474, -105.194891)[1].

According to the United States Census Bureau, the town has a total area of 0.3 square miles (0.7 km²), all of it land.

Demographics

As of the census[2] of 2000, there were 911 people, 325 households, and 206 families residing in the town. The population density was 3,406.0 people per square mile (1,302.7/km²). There were 350 housing units at an average density of 1,308.5/sq mi (500.5/km²). The racial makeup of the town was 32.16% White, 0.11% African American, 63.67% Native American, 0.77% Asian, 0.22% from other races, and 3.07% from two or more races. Hispanic or Latino of any race were 0.88% of the population.

There were 325 households out of which 33.5% had children under the age of 18 living with them, 36.9% were married couples living together, 19.4% had a female householder with no husband present, and 36.6% were non-families. 32.3% of all households were made up of individuals and 12.6% had someone living alone who was 65 years of age or older. The average household size was 2.57 and the average family size was 3.26.

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