1900 United States presidential election in Washington (state)

1900 United States presidential election in Washington (state)

← 1896 November 6, 1900 1904 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai Stevenson I
Electoral vote 4 0
Popular vote 57,456 44,833
Percentage 53.44% 41.70%

County Results

President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in Washington took place on November 6, 1900. All contemporary 45 states were part of the 1900 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.

Washington was won by the Republican nominees, incumbent President William McKinley of Ohio and his running mate Theodore Roosevelt of New York. They defeated the Democratic nominees, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan and his running mate, former Vice President Adlai Stevenson I. McKinley won the state by a margin of 11.74% in this rematch of the 1896 presidential election. The return of economic prosperity, recent victory in the Spanish–American War, continued American expansion in the Philippines,[1] and the fading of the Populist revolt of the previous decade ensured that incumbent President McKinley would not have any trouble carrying the state.

McKinley had previously lost Washington to Bryan four years earlier while Bryan would later lose the state to William Howard Taft in 1908.

  1. ^ Gates, John M.; ‘Philippine Guerrillas, American Anti-Imperialists, and the Election of 1900’, Pacific Historical Review, vol. 46, no. 1 (February 1977), pp. 51-64

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