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1868 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1868 United States presidential election in Rhode Island

← 1864 November 3, 1868 1872 →
 
Ulysses S Grant by Brady c1870-restored (cropped).jpg
Horatio Seymour - Brady-Handysmall.jpg
Nominee Ulysses S. Grant Horatio Seymour
Party Republican Democratic
Home state Illinois New York
Running mate Schuyler Colfax Francis Preston Blair, Jr.
Electoral vote 4 0
Popular vote 12,993 6,548
Percentage 66.49% 33.51%

President before election

Andrew Johnson
Democratic

Elected President

Ulysses S. Grant
Republican

The 1868 United States presidential election in Rhode Island took place on November 3, 1868, as part of the 1868 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 Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 32.98%.

With 66.49% of the popular vote, Rhode Island would be Grant's fifth strongest victory in terms of popular vote percentage after Vermont, Massachusetts, Kansas and Tennessee.[1]

Results

1868 United States presidential election in Rhode Island[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 12,993 66.49% 4 100.00%
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 6,548 33.51% 0 0.00%
Total 19,541 100.00% 4 100.00%


References

  1. ^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1868 Presidential General Election Results - Rhode Island".
This page was last edited on 11 July 2019, at 18:45
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