Election in Alabama
1884 United States presidential election in Alabama
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Turnout | 12.03% of the total population 0.14 pp[1] |
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| | | Nominee | Grover Cleveland | James G. Blaine | | Party | Democratic | Republican | Home state | New York | Maine | Running mate | Thomas A. Hendricks | John A. Logan | Electoral vote | 10 | 0 | Popular vote | 92,736 | 59,444 | Percentage | 60.37% | 38.69% | |
County results Cleveland 50–60% 60–70% 70–80% 80–90% 90–100% | Blaine 40–50% 50–60% 60–70% | |
President before election James A. Garfield Republican | Elected President Grover Cleveland Democratic | |
Elections in Alabama |
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The 1884 United States presidential election in Alabama took place on November 4, 1884, as part of the nationwide presidential election. Alabama voters chose ten representatives, or electors, to the Electoral College, who voted for president and vice president.[2]
Alabama was won by Grover Cleveland, the 28th governor of New York, (D–New York), running with the former governor of Indiana Thomas A. Hendricks, with 60.37% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 38.69% of the vote.[2]
Results
See also
References
- ^ "1884 Presidential Election Results Alabama Total Population Turnout".
- ^ a b c "1884 Presidential Election Results Alabama".
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