July 26, 2020 – 6:07 PM
Florida’s 423,855 COVID-19 cases through Sunday were only surpassed by California’s 453,659 cases
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Florida overtook New York State over the weekend as the state with the second-highest number of cases of COVID-19 in the US, as more than 9,300 new cases were reported on Sunday in the Sunshine State, accompanied by 78 additional new deaths.
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The 423,855 cases of COVID-19 Florida’s as of Sunday were surpassed only by 453,659 California cases. With 39.5 million residents, California has nearly twice the population of Florida’s 21.4 million. California is the most populous state in the nation, followed by Texas, Florida, and New York.
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New York, once the epicenter of the virus in the United States, had 411,736 cases of coronavirus. The state has 19.4 million residents.
There were 9,344 new cases reported in Florida on Sunday. The number of new cases was lower than other days last week, but the cases launched on Sundays tend to be lower due to a lack of workers entering data or in laboratories analyzing samples.
The median statewide age of coronavirus patients in Florida was 40.
Nearly 3.4 million Floridians have been screened for viruses. New cases tested over the weekend had a positive rate of 11%, according to the Florida Department of Health.
The state health department recorded 781 deaths from COVID-19 over the past week for an average of 126 deaths per day on Sunday, slightly below Saturday’s weekly average of 127 deaths per day. Florida had 5,972 total deaths as of Sunday, according to the Florida Department of Health.
Florida also had 8,951 coronavirus-related hospitalizations as of Sunday.
– SOURCE: AP–
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