1900 United States presidential election in Wyoming information
Election in Wyoming
Main article: 1900 United States presidential election
1900 United States presidential election in Wyoming
← 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
3
0
Popular vote
14,482
10,164
Percentage
58.66%
41.17%
County Results
McKinley
40-50%
50-60%
60-70%
President before election
William McKinley
Republican
Elected President
William McKinley
Republican
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v
t
e
The 1900 United States presidential election in Wyoming took place on November 6, 1900, as part of the 1900 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president.
Wyoming was won by the President William McKinley (R–Ohio), running with the 33rd Governor of New York, Theodore Roosevelt, with 58.66 percent of the popular vote, against representative William Jennings Bryan (D–Nebraska), running with the 23rd Vice President Adlai Stevenson I, with 41.17 percent of the popular vote.[1] McKinley won the state by a margin of 17.49%
McKinley had previously lost Wyoming to Bryan four years earlier while Bryan would later go on to lose the state again to William Howard Taft in 1908.
^"1900 Presidential Election Results Wyoming".
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