http://medsocial.com/blog.aspx?blogaction=viewblog&show=292
Story that says so
2007-03-08 13:09:06 · 12 answers · asked by Adam G 1 in Health ➔ General Health Care ➔ Injuries
They always seem at half-staff on weekends- I'm sure the death rate goes up in the evenings as well.
2007-03-08 13:12:20 · answer #1 · answered by pavano_carl 4 · 1⤊ 1⤋
Several factors may play into this. People are more apt to be out drinking and driving on weekends. Families tend to be with dying patients on weekends and may decide to "pull the plug" as a group. More people have heart attacks when at home doing things not in their usual routine (cutting the grass, shoveling snow, etc). I don't believe there is diminshed care on weekends.
2007-03-08 13:15:26 · answer #2 · answered by Dino4747 5 · 1⤊ 1⤋
I am not sure- you'd probably have to gather alot of data and explore that assumption. I don't think doctors and nurses and other hosptial staff are any less responsible on a saturday than on a monday- if so that hospital needs to be investigated. if this was the case- it'd be another sensationalized story as well.
2007-03-08 13:11:28 · answer #3 · answered by actuatedtendancy 2 · 0⤊ 1⤋
I would believe that. People are more active on the weekends and do more items that would put them at a higher risk.
2007-03-08 13:18:40 · answer #4 · answered by Mariposa 7 · 1⤊ 1⤋
yes nice story
2007-03-08 13:11:50 · answer #5 · answered by dodgedifferent01 3 · 0⤊ 1⤋
doctors and nurses are more laid back on weekends?
2007-03-08 13:10:48 · answer #6 · answered by UnK3 2 · 0⤊ 1⤋
if the story said so then why did you just give me 2 POINTS!!! YEAH!!!!!!!!!!
2007-03-08 13:11:06 · answer #7 · answered by Anonymous · 0⤊ 1⤋
YES, I BELIEVE THEY DO
2007-03-08 13:16:46 · answer #8 · answered by Swm 39 4 Younger Swf Forever 4 · 0⤊ 1⤋
oh..............this question is dumb...................YOU DIE WHEN ITS YOUR TIME................ITS ALL IN GODS HANDS
2007-03-08 13:15:21 · answer #9 · answered by ice 2 · 0⤊ 0⤋
yep.
2007-03-08 13:11:56 · answer #10 · answered by Anonymous · 0⤊ 1⤋