1900 United States presidential election in Rhode Island

1900 United States presidential election in Rhode Island

← 1896 November 6, 1900 1904 →

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Theodore Roosevelt Adlai E. Stevenson Electoral vote 4 0 Popular vote 33,784 19,812 Percentage 59.74% 35.04%

County Results McKinley   50-60%   60-70%

President before election William McKinley Republican Elected President William McKinley Republican

Elections in Rhode Island
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Ballot measures 2020 Question 1
Providence Mayoral elections 1998 2002 2006 2010 2014 2018 2022
v t e

The 1900 United States presidential election in Rhode Island took place on November 6, 1900, as part of the 1900 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won Rhode Island by a margin of 24.7% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

Bryan had previous lost Rhode Island to McKinley four years earlier and would later lose the state again in 1908 to William Howard Taft.