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1976 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1976 United States presidential election in Rhode Island

← 1972 November 2, 1976 1980 →
Jimmy Carter (cropped).jpg
Gerald Ford, official Presidential photo.jpg
Nominee Jimmy Carter Gerald Ford
Party Democratic Republican
Home state Georgia Michigan
Running mate Walter Mondale Bob Dole
Electoral vote 4 0
Popular vote 227,636 181,249
Percentage 55.4% 44.1%

Rhode Island Presidential Election Results 1976.svg
County Results

President before election

Gerald Ford

Elected President

Jimmy Carter

The 1976 United States presidential election in Rhode Island took place on November 2, 1976, as part of the 1976 United States presidential election. Voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

Rhode Island was won by Jimmy Carter (DGeorgia), with 55.36% of the popular vote. Carter defeated incumbent President Gerald Ford (RMichigan), who finished with 44.08% of the popular vote. No third-party candidate received any votes.

Jimmy Carter went on to become the 39th president of the United States.[2]


1976 United States presidential election in Rhode Island
Party Candidate Votes %
Democratic Jimmy Carter 227,636 55.36%
Republican Gerald Ford (inc.) 181,249 44.08%
Write-In 2,285 0.56%
Total votes 411,170 100%

By county

County Carter% Carter# Ford% Ford# Others% Others#
Providence 58% 144,805 41.6% 103,976 0.4% 1,036
Newport 53.7% 17,768 45.8% 15,155 0.6% 184
Bristol 52.4% 11,228 47.3% 10,131 0.3% 66
Kent 51.1% 35,855 48.6% 34,131 0.3% 227
Washington 49.9% 17,980 49.6% 17,856 0.5% 186

See also


  1. ^ This figure is calculated by dividing the total number of votes cast in 1976 (416,583) by an estimate of the number of registered voters in Rhode Island in 1976 (544,992). See "General Election November 2, 1976". Rhode Island Board of Elections. Retrieved February 6, 2018.
  2. ^

This page was last edited on 9 March 2021, at 13:54
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