United States presidential election in Iowa, 1944
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All 10 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
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The 1944 United States presidential election in Iowa took place on November 7, 1944, as part of the 1944 United States presidential election. Iowa voters chose ten[2] representatives, or electors, to the Electoral College, who voted for president and vice president.
Iowa was won by Governor Thomas E. Dewey (R–New York), running with Governor John Bricker, with 51.99% of the popular vote, against incumbent President Franklin D. Roosevelt (D–New York), running with Senator Harry S. Truman, with 47.49% of the popular vote.[3][4]
Results
Party | Candidate | Votes | % | |
---|---|---|---|---|
Republican | Thomas E. Dewey | 547,267 | 51.99% | |
Democratic | Franklin D. Roosevelt (inc.) | 499,876 | 47.49% | |
Prohibition | Claude A. Watson | 3,752 | 0.36% | |
Write-in | 1,704 | 0.16% | ||
Total votes | 1,052,599 | 100% |
References
- ↑ "United States Presidential election of 1944 - Encyclopædia Britannica". Retrieved December 22, 2017.
- ↑ "1944 Election for the Fortieth Term (1945-49)". Retrieved December 22, 2017.
- ↑ "1944 Presidential General Election Results - Iowa". Retrieved December 22, 2017.
- ↑ "The American Presidency Project - Election of 1944". Retrieved December 22, 2017.
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