Delta flight DL6408 schedule

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1. Flight DL6408: Cleveland - New York City, CLE - JFK, 1h 49m

Cleveland, Cleveland-Hopkins (CLE)
16:21
(4:21 pm)
23 Aug
 
Delta
DL6408
1h 49m
New York City, John F. Kennedy (JFK)
18:10
(6:10 pm)
23 Aug
Terminal 2
Delta domestic flight DL6408 departs from airport Cleveland, Cleveland-Hopkins (CLE), United States on Saturday, 23 August at 16:21 / 4:21 pm. The flight arrives to «Terminal 2» of airport New York City, John F. Kennedy (JFK), United States on Saturday, 23 August at 18:10 / 6:10 pm. Flight duration is 1h 49m.
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