US holiday weekend adds to coronavirus worries as Florida reports 11,445 new cases

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Florida reported a record number of coronavirus cases on Saturday, the latest sign that the virus is surging in many parts of the United States, casting a pall over Fourth of July celebrations.

Officials and health authorities warned people to take precautions or simply stay home on Independence Day, as confirmed cases are climbing in 40 states. The US set another daily record Friday with 52,300 newly reported infections, according to numbers kept by Johns Hopkins University.

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The US has more than 2.8 million confirmed cases — about a quarter of worldwide infections, according to the tally, which is widely thought to understate the true toll, partially because of asymptomatic cases and limited testing.

Florida reported 11,445 confirmed infections on Saturday, bringing the statewide total to more than 190,000.

A beachgoer dances on the pier at Coney Island on the Fourth of July holiday in Brooklyn, New York City. (Reuters)
A beachgoer dances on the pier at Coney Island on the Fourth of July holiday in Brooklyn, New York City. (Reuters)

Alabama this week also set a record for infections reported in a day, with more than 1,700 confirmed cases recorded Thursday, and state health officials worried that the holiday weekend could lead to the type of spike that followed Memorial Day weekend.

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“I’m really, really worried about the Fourth of July,” said Dr. Don Williamson, a former state health officer who now heads the Alabama Hospital Association. “I think that will likely determine the trend for Alabama for the rest of the summer.”

The sobering updates came as local officials and health experts across the country tried to minimize opportunities for the virus to continue spreading on a holiday weekend typically spent swimming, eating at backyard gatherings or crowding elbow-to-elbow to view parades and fireworks shows.

Many communities canceled those events and cautioned people against gathering on their own.

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