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American Airlines AA7588 Flight Status Today
International flight American Airlines AA7588 takes off from Belize City (BZE) Belize to Los Angeles (LAX) United States. It's operated by American Airlines. The plane leaves Philip S.W. Goldson International Airport at 11:08 31 December 2025 America/Belize. The flight is expected to land at Los Angeles International Airport at 18:19 31 December 2025 America/Los_Angeles. The flight will last about 3 hours 56 minutes.
Current status of AA7588 is Arrived On time
- Type: International Flight
- Code Shares: HA 6740
- Flight Duration: 3 hours 56 minutes
- Flight Distance: 3504 kms / 2177 miles
- American Airlines
- IATA: AA
- ICAO: AAL
- Operating Days: Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
- Service type: Normal passenger
- Seats: 76
- Freight capacity: 3 tons
- Passenger classes: Economy, Business Class
- Aircraft: Boeing 737-900 Passenger (Scimitar Winglets)
- Callsign: AAL7588
- Departure Timezone: America/Belize
- Arrival Timezone: America/Los_Angeles
- Current Time in Belize City: Wednesday 2025-12-31 23:35 PM
- Current Time in Los Angeles: Wednesday 2025-12-31 21:35 PM
Flight routes similar to American Airlines AA7588
American Airlines Flight AA7588 FAQs
What is the scheduled flight duration for American Airlines AA7588 flight?
On average, nonstop flight takes 3 hour(s) 56 minutes, with the flight distance of 3504 km (2177 miles).
What is the current status of American Airlines AA7588 flight?
The current status of flight American Airlines AA7588 is Arrived On time.
What type of aircraft is used for the American Airlines AA7588 flight?
All American Airlines AA7588 flights are operated using Boeing 737-900 Passenger (Scimitar Winglets) aircraft.
How many American Airlines AA7588 flights are operated a week?
6 flights per week. The Flight American Airlines AA7588 is operated on Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
