[timeout:300][out:json]; ( node(around:1000,28.07361,-80.68583)[boundary=administrative][~"^(addr:housenumber|.*name.*)$"~".",i]; way(around:1000,28.07361,-80.68583)[boundary=administrative][~"^(addr:housenumber|.*name.*)$"~".",i]; rel(around:1000,28.07361,-80.68583)[boundary=administrative][~"^(addr:housenumber|.*name.*)$"~".",i]; node(around:1000,28.07361,-80.68583)[boundary=census][~"^(addr:housenumber|.*name.*)$"~".",i]; way(around:1000,28.07361,-80.68583)[boundary=census][~"^(addr:housenumber|.*name.*)$"~".",i]; rel(around:1000,28.07361,-80.68583)[boundary=census][~"^(addr:housenumber|.*name.*)$"~".",i]; ); out center tags;
June Park is a census-designated place (CDP) in Brevard County, Florida. The population was 4,283 at the 2020 United States census, up from 4,094 at the 2010 United States Census. It is part of the Palm Bay–Melbourne–Titusville, Florida Metropolitan Statistical Area.
June Park es un lugar designado por el censo ubicado en el condado de Brevard en el estado estadounidense de Florida. En el Censo de 2010 tenía una población de 4.094 habitantes y una densidad poblacional de 465,6 personas por km².[2]
found 2 match candidates
census-designated place in the United States (Q498162) | boundary=census |
administrative territorial entity (Q56061) | boundary=administrative |