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1960 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

United States presidential election in Rhode Island, 1960

← 1956 November 8, 1960 1964 →
 
John F. Kennedy, White House color photo portrait.jpg
VP-Nixon.png
Nominee John F. Kennedy Richard Nixon
Party Democratic Republican
Home state Massachusetts California
Running mate Lyndon B. Johnson Henry Cabot Lodge, Jr.
Electoral vote 4 0
Popular vote 258,032 147,502
Percentage 63.63% 36.36%

President before election

Dwight Eisenhower
Republican

Elected President

John F. Kennedy
Democratic

The 1960 United States presidential election in Rhode Island took place on November 8, 1960, as part of the 1960 United States presidential election, which was held throughout all 50 states. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Democratic nominee, Senator John F. Kennedy of Massachusetts, over the Republican nominee, Vice President Richard Nixon of California. Kennedy ran with Senate Majority Leader Lyndon B. Johnson of Texas, while Nixon's running mate was Ambassador Henry Cabot Lodge, Jr. of Massachusetts.

Kennedy carried Rhode Island by a margin of 27.27 percent.

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Transcription

Results

United States presidential election in Rhode Island, 1960[1]
Party Candidate Votes Percentage Electoral votes
Democratic John F. Kennedy 258,032 63.63% 4
Republican Richard Nixon 147,502 36.36% 0
Write-ins Write-ins 1 0.01% 0
Totals 405,535 100.00% 4

By county

County Kennedy% Kennedy# Nixon% Nixon# Others% Others#
Providence 67.5% 189,014 32.5% 91,028 0% 0
Bristol 59.6% 11,099 40.4% 7,537 0% 0
Newport 56.8% 15,677 43.2% 11,942 0% 0
Kent 55.7% 30,662 44.3% 24,344 0% 0
Washington 47.8% 11,580 52.2% 12,651 0% 0

References

  1. ^ "1960 Presidential General Election Results - Rhode Island". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved 2013-02-07.


This page was last edited on 6 May 2019, at 19:30
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