Belgrade, Nebraska

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Belgrade is a village in Nance County, Nebraska, United States. The population was 134 at the 2000 census.

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[edit] Geography

Location of Belgrade, Nebraska

Belgrade is located at 41°28′17″N, 98°4′2″W (41.471327, -98.067302)GR1.

According to the United States Census Bureau, the village has a total area of 0.5 km² (0.2 mi²), all land.

[edit] Demographics

As of the censusGR2 of 2000, there were 134 people, 63 households, and 36 families residing in the village. The population density was 287.4/km² (730.3/mi²). There were 77 housing units at an average density of 165.2/km² (419.6/mi²). The racial makeup of the village was 100.00% White.

There were 63 households out of which 23.8% had children under the age of 18 living with them, 47.6% were married couples living together, 4.8% had a female householder with no husband present, and 41.3% were non-families. 38.1% of all households were made up of individuals and 15.9% had someone living alone who was 65 years of age or older. The average household size was 2.13 and the average family size was 2.73.

In the village the population was spread out with 23.9% under the age of 18, 6.7% from 18 to 24, 22.4% from 25 to 44, 28.4% from 45 to 64, and 18.7% who were 65 years of age or older. The median age was 42 years. For every 100 females there were 94.2 males. For every 100 females age 18 and over, there were 92.5 males.

The median income for a household in the village was $28,750, and the median income for a family was $32,143. Males had a median income of $28,750 versus $21,875 for females. The per capita income for the village was $12,767. There were 12.5% of families and 15.8% of the population living below the poverty line, including no under eighteens and 33.3% of those over 64.

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