1900 United States presidential election in Rhode Island

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.

1900 United States presidential election in Rhode Island

November 6, 1900

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

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.

Results

1900 United States presidential election in Rhode Island[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	William McKinley of Ohio	Theodore Roosevelt of New York	33,784	59.74%	4	100.00%
	Democratic	William Jennings Bryan of Nebraska	Adlai Ewing Stevenson I of Illinois	19,812	35.04%	0	0.00%
	Prohibition	John Granville Woolley of Illinois	Henry Brewer Metcalf of Rhode Island	1,529	2.70%	0	0.00%
	Socialist Labor	Joseph F. Malloney of Massachusetts	Valentine Remmel of Pennsylvania	1,423	2.52%	0	0.00%
Total				56,198	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island

References

1. "1900 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.