1908_United_States_presidential_election_in_Rhode_Island

1908 United States presidential election in Rhode Island

1908 United States presidential election in Rhode Island

Election in Rhode Island

The 1908 United States presidential election in Rhode Island took place on November 3, 1908 as part of the 1908 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Main article: 1908 United States presidential election
Quick Facts Nominee, Party ...

Rhode Island voted for the Republican nominees, Secretary of War William Howard Taft of Ohio and his running mate James S. Sherman of New York. They defeated the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate John W. Kern of Indiana. Taft won the state by a margin of 26.6%.

With 60.76% of the popular vote, Rhode Island would be Taft's fifth strongest victory in terms of percentage in the popular vote after Vermont, Maine, Michigan and North Dakota.[1]

Bryan had previously lost Rhode Island to William McKinley in both 1896 and 1900.

Results

More information Party, Candidate ...

See also

References

  1. [1]
    "1908 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. [2]
    "1908 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.
Share this article:

This article uses material from the Wikipedia article 1908_United_States_presidential_election_in_Rhode_Island, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.