Alfred P. Swineford
(1836-1909)

American politician (1836-1909) – Alfred P. Swineford was born in Ashland (City in Ohio, United States) on September 14th, 1836 and died in Juneau (State capital of Alaska, United States) on October 26th, 1909 at the age of 73. Today Alfred P. Swineford would be 187 years old.

Age

how old was Alfred P. Swineford when he died?

73

Alfred P. Swineford died in 1909 at the age of 73.


Biographical data

Birthday September 14th, 1836 (Wednesday)
Place of Birth Ashland
City in Ohio, United States
Death Date October 26th, 1909 (Tuesday)
Death place Juneau
State capital of Alaska, United States
Birth sign (Zodiac) Virgo (The Maiden) ♍
Chinese Zodiac Monkey 猴

Other personalities born on September 14

Also born in 1836

Alfred P. Swineford

Birthday

When was Alfred P. Swineford born?

Alfred P. Swineford was born on September 14th, 1836.

Death

When did Alfred P. Swineford die?

Alfred P. Swineford died on October 26th, 1909 at the age of 73 in Juneau (State capital of Alaska, United States). Today Alfred P. Swineford would be 187 years old.

Place of Birth

Where was Alfred P. Swineford born?

Ashland (City in Ohio, United States).

Death place

Where did Alfred P. Swineford die?

Juneau (State capital of Alaska, United States).

Zodiac

What is the zodiac sign of Alfred P. Swineford?

Alfred P. Swineford was born in the zodiac sign Virgo (The Maiden).

Chinese Zodiac

Which Chinese zodiac sign does Alfred P. Swineford have?

Alfred P. Swineford was born in 1836 in the year of the Chinese zodiac Monkey.

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Died on October 26th, 1909:
Anniversaries of death

American politician (1836-1909)

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