1892 United States presidential election in Minnesota
The 1892 United States presidential election in Minnesota took place on November 8, 1892. All contemporary 44 states were part of the 1892 United States presidential election. Minnesota voters chose nine electors to the Electoral College, which selected the president and vice president.
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Elections in Minnesota | ||||||||||
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State executive elections
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Minnesota was won by the Republican nominees, incumbent President Benjamin Harrison of Indiana and his running mate Whitelaw Reid of New York.
With 5.31% of the popular vote, Minnesota proved to be Bidwell’s best performance as well as the only state where he broke 5% of the vote.[1]
Results
1892 United States presidential election in Minnesota[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | Benjamin Harrison (incumbent) | 122,823 | 45.96% | 9 | |
Democratic | Grover Cleveland | 100,920 | 37.76% | 0 | |
People's | James Weaver | 29,313 | 10.97% | 0 | |
Prohibition | John Bidwell | 14,182 | 5.31% | 0 | |
Totals | 267,238 | 100.00% | 9 | ||
Voter turnout | — |
gollark: Yes.
gollark: You could use out of band signalling the same way the beeoid replication works.
gollark: I mean, yours is basically tit for tat, yes?
gollark: That way you lose 0 points (except in the initial signalling sequence?) but *other* people will also lose lots.
gollark: THEY need to cooperate.
References
- "1892 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- Dave Leip's U.S. Election Atlas; Presidential General Election Results – Minnesota
Notes
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