1948 United States presidential election in Rhode Island
The 1948 United States presidential election in Rhode Island took place on November 7, 1944, as part of the 1948 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
 County Results
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Elections in Rhode Island |
|---|
 |
Rhode Island was won by Democratic candidate, incumbent President Harry S. Truman over Republican candidate New York governor Thomas E. Dewey.
Truman won Rhode Island by a margin of 16.15%.
Results
| 1948 United States presidential election in Rhode Island[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Harry S. Truman of Missouri | Alben William Barkley of Kentucky | 188,736 | 57.59% | 4 | 100.00% | ||
| Republican | Thomas Edmund Dewey of New York | Earl Warren of California | 135,787 | 41.44% | 0 | 0.00% | ||
| Progressive | Henry Agard Wallace of Iowa | Glen Hearst Taylor of Idaho | 2,619 | 0.80% | 0 | 0.00% | ||
| Socialist | Norman Thomas of New York | Tucker Powell Smith of Michigan | 429 | 0.13% | 0 | 0.00% | ||
| Socialist Labor | Edward A. Teichert of Pennsylvania | Stephen Emery of New York | 131 | 0.04% | 0 | 0.00% | ||
| Total | 327,702 | 100.00% | 4 | 100.00% | ||||
References
- "1948 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
| |
|
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.