Английская Википедия:1864 United States presidential election in Rhode Island
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Шаблон:Short description Шаблон:Use mdy dates Шаблон:Main Шаблон:Infobox election Шаблон:Elections in Rhode Island sidebar The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the National Union candidate, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%.
Results
1864 United States presidential election in Rhode Island[1] | |||||
---|---|---|---|---|---|
Party | Candidate | Votes | Percentage | Electoral votes | |
National Union | Abraham Lincoln (incumbent) | 13,962 | 62.24% | 4 | |
Democratic | George B. McClellan | 8,470 | 37.76% | 0 | |
Totals | 22,432 | 100.0% | 4 |
See also
References
Шаблон:State Results of the 1864 U.S. presidential election
Шаблон:RhodeIsland-election-stub