1968 United States presidential election in Rhode Island

1968 United States presidential election in Rhode Island

← 1964 November 5, 1968[1] 1972 →
 
Nominee Hubert Humphrey Richard Nixon
Party Democratic Republican
Home state Minnesota New York[a]
Running mate Edmund Muskie Spiro Agnew
Electoral vote 4 0
Popular vote 246,518 122,359
Percentage 64.03% 31.78%


President before election

Lyndon B. Johnson
Democratic

Elected President

Richard Nixon
Republican

The 1968 United States presidential election in Rhode Island took place on November 5, 1968, as part of the 1968 United States presidential election. Voters chose four[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Rhode Island was won by the Democratic candidate, Vice President Hubert Humphrey, with 64.03% of the popular vote, against the Republican candidate, former Senator and Vice President Richard Nixon, with 31.78% of the popular vote. American Independent Party candidate George Wallace also appeared on the ballot, finishing with 4.07% of the popular vote.[3][4] Despite the state trending 30 points Republican after setting a record in 1964, the state continues to overwhelmingly vote for the Democratic candidate.

Nixon was the first Republican to win the presidency without gaining the majority vote in Washington, Kent, and Newport counties.

  1. ^ "United States Presidential election of 1968 - Encyclopædia Britannica". Retrieved May 24, 2017.
  2. ^ "1968 Election for the Forty-Sixth Term (1969-73)". Retrieved May 24, 2017.
  3. ^ "1968 Presidential General Election Results - Rhode Island". Retrieved May 24, 2017.
  4. ^ "The American Presidency Project - Election of 1968". Retrieved May 24, 2017.


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