1892 United States presidential election in Delaware
The 1892 United States presidential election in Delaware took place on November 8, 1892. All contemporary 44 states were part of the 1892 United States presidential election. Delaware voters chose 3 electors to the Electoral College, which selected the president and vice president.
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Delaware was won by the Democratic nominees, former President Grover Cleveland of New York and his running mate Adlai Stevenson I of Illinois.
Results
United States presidential election in Delaware, 1892[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Grover Cleveland | 18,581 | 49.90% | 3 | |
Republican | Benjamin Harrison (incumbent) | 18,077 | 48.55% | 0 | |
Prohibition | John Bidwell | 564 | 1.51% | 0 | |
Write-ins | Scattered | 13 | 0.03% | 0 | |
Totals | 37,235 | 100.00% | 3 | ||
Voter turnout | — |
gollark: Troubling.
gollark: I can *read* those numbers.
gollark: How come APL doesn't have some accursed high-density number encoding?
gollark: Perhaps these things are correlated with politics. Someone should check.
gollark: You mean 8values?
References
- Dave Leip's U.S. Election Atlas; Presidential General Election Results – Delaware
Notes
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