United States presidential election in Maryland, 1892
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The 1892 United States presidential election in Maryland took place on November 8, 1892. All contemporary 44 states were part of the 1892 United States presidential election. Maryland voters chose eight electors to the Electoral College, which selected the president and vice president.
Maryland 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 Maryland, 1892[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Grover Cleveland | 113,866 | 53.39% | 8 | |
Republican | Benjamin Harrison (incumbent) | 92,736 | 43.48% | 0 | |
Prohibition | John Bidwell | 5,877 | 2.76% | 0 | |
People's | James Weaver | 796 | 0.37% | 0 | |
Totals | 213,275 | 100.00% | 8 | ||
Voter turnout | — |
References
- ↑ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Maryland
Notes
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