Three passengers on the same Delta Airlines flight tested positive for the coronavirus after arriving in New York, officials announced
The three individuals began feeling symptoms on July 6, the day after traveling from Atlanta to Albany on Delta Flight 4815, Rensselaer County Health Department officials said in a statement.
“Three passengers, all from Rensselaer County, became symptomatic on July 7. The cases were confirmed to the county Health Department today, Friday, July 10, after testing,” county officials wrote on Facebook Friday.
The airline is now warning any other passengers on the flight from Georgia to “immediately contact the county Health Department in their home county.”
Delta received $5.4 billion in bailout money from the federal government earlier this year, MarketWatch reported. The company’s website says it plans to block middle seats and limit capacity on its planes until Sept. 30.
The airline also requires all passengers to wear masks.
A Delta rep said the agency provided local health officials with a copy of the flight’s manifest for contact tracing purposes.
“We have been made aware of three customers who tested positive for Covid-19 and recently traveled on Endeavor from Atlanta to Albany,” spokesman Anthony Black told CNN.
“We are following the guidance of local healthcare officials and the Centers for Disease Control and Prevention. The health and safety of our customers and crews is our top priority.”
New York state requires travelers from 19 states with rising coronavirus rates — including Georgia — to self-quarantine for 14 days.
On Monday, Gov. Andrew Cuomo said air travelers from those states must fill out identification forms upon arrival in New York — or face a $2,000 fine.
“Out-of-state travelers from the states that are quarantined must provide a location form before they leave the airport,” the governor said.
“We know there’s been instances of noncompliance. Noncompliance can lead to outbreaks.”