American Hall of Fame baseball player (1879-1930) – Rube Foster was born in Calvert (City in Robertson County, Texas, United States) on September 17th, 1879 and died in Kankakee (City in Kankakee County, Illinois, United States) on December 9th, 1930 at the age of 51. Today Rube Foster would be 144 years old.
Rube Foster died in 1930 at the age of 51.
Birthday | September 17th, 1879 (Wednesday) |
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Place of Birth | Calvert City in Robertson County, Texas, United States |
Death Date | December 9th, 1930 (Tuesday) |
Death place | Kankakee City in Kankakee County, Illinois, United States |
Birth sign (Zodiac) | Virgo (The Maiden) ♍ |
Chinese Zodiac | Rabbit 兔 |
Rube Foster was born on September 17th, 1879.
Rube Foster died on December 9th, 1930 at the age of 51 in Kankakee (City in Kankakee County, Illinois, United States). Today Rube Foster would be 144 years old.
Calvert (City in Robertson County, Texas, United States).
Kankakee (City in Kankakee County, Illinois, United States).
Rube Foster was born in the zodiac sign Virgo (The Maiden).
Rube Foster was born in 1879 in the year of the Chinese zodiac Rabbit.