1996_United_States_presidential_election_in_Rhode_Island

1996 United States presidential election in Rhode Island

1996 United States presidential election in Rhode Island

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The 1996 United States presidential election in Rhode Island took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Quick Facts Turnout, Nominee ...

Rhode Island was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 59.71% to 26.82% by a margin of 32.89%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third, with 11.20% of the popular vote.[2]

As of 2020, this was the most recent presidential election in which the town of Scituate voted for a Democrat.

Results

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See also


References

  1. This figure is calculated by dividing the total number of votes cast in 1996 (390,247) by an estimate of the number of registered voters in Rhode Island in 1996 (602,692). See "Voter Turnout, 1996". Rhode Island Board of Elections. Retrieved February 6, 2018.

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