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1912 United States presidential election in Rhode Island

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1912 United States presidential election in Rhode Island
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The 1912 United States presidential election in Rhode Island took place on November 5, 1912, as part of the 1912 United States presidential election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

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Rhode Island was won by the Democratic Party nominees, New Jersey Governor Woodrow Wilson and Indiana Governor Thomas R. Marshall. Wilson and Marshall defeated incumbent President William Howard Taft, and his running mate Vice President James S. Sherman and Progressive Party candidates, former President Theodore Roosevelt and his running mate California Governor Hiram Johnson.

Wilson won Rhode Island by a narrow margin of 3.48 points, becoming the first Democratic presidential candidate since Franklin Pierce in 1852 to win the state. Wilson also became the first Democrat since Pierce to win Providence County or any county in the state. Another Democratic candidate wouldn't win Rhode Island again until Al Smith won it in 1928. This was the first of three times that the state voted differently than Minnesota, along with 1928 and 1984.

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Results

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See also

Notes

  1. In this town where Roosevelt ran ahead of Wilson, margin given is Taft vote minus Roosevelt vote and percentage margin is Taft percentage minus Roosevelt percentage.
  2. In this town where Roosevelt ran ahead of Taft, margin given is Wilson vote minus Roosevelt vote and percentage margin is Wilson percentage minus Roosevelt percentage.

References

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