1884 United States presidential election in Rhode Island

Main article: 1884 United States presidential election
1884 United States presidential election in Rhode Island
November 4, 1884
 
Nominee James G. Blaine Grover Cleveland
Party Republican Democratic
Home state Maine New York
Running mate John A. Logan Thomas A. Hendricks
Electoral vote 4 0
Popular vote 19,030 12,391
Percentage 58.07% 37.81%
President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in Rhode Island occurred on November 4, 1884. This election was part of the 1884 United States presidential election. Voters in Rhode Island selected four representatives, or electors, to the Electoral College, which elects the president and vice president.

Rhode Island voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 20.26%.

With 58.07% of the popular vote, Rhode Island became Blaine's fourth strongest victory in terms of percentage, following Vermont, Minnesota and Kansas.

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