1944 United States presidential election in Rhode Island

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United States presidential election in Rhode Island, 1944

← 1940 November 7, 1944 1948 →
  1944 portrait of FDR (1)(small).jpg ThomasDewey.png
Nominee Franklin D. Roosevelt Thomas E. Dewey
Party Democratic Republican
Home state New York New York
Running mate Harry S. Truman John W. Bricker
Electoral vote 4 0
Popular vote 175,356 123,487
Percentage 58.59% 41.26%

President before election

Franklin D. Roosevelt
Democratic

Elected President

Franklin D. Roosevelt
Democratic

The 1944 United States presidential election in Rhode Island took place on November 7, 1944, as part of the 1944 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.

Rhode Island was won by Democratic candidate, incumbent President Franklin D. Roosevelt won the state over the Republican candidate New York governor Thomas E. Dewey.

Roosevelt won the state by a margin of 17.33 percent.

Results[edit]

United States presidential election in Rhode Island, 1944[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Delano Roosevelt of New York Harry S. Truman of Missouri 175,356 58.59% 4 100.00%
Republican Thomas Edmund Dewey of New York John William Bricker of Ohio 123,487 41.26% 0 0.00%
Prohibition Claude A. Watson of California Andrew Nathan Johnson of Kentucky 433 0.14% 0 0.00%
Total 299,276 100.00% 4 100.00%

References[edit]

  1. ^ "1944 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.