From what I can gather, the average suvivability of covid is commonly known as 99.7% of all age groups. Can you break it down for me on how they are calculating 99.7% from the official numbers of the cdc?
Thanks.
Edit: Is this correct, deaths/cases x 100?
From cdc, currently, 37,259,886 cases 623,244 deaths.
But I am getting 1.6%, that's 98.33% (maybe 99.7% was calc from previous numbers if someone knows?).
Sidenote, we all know the actual death is 5.5% of the covid deaths they report.
See here for one study, placing the mortality rate as ~0.15%, averaged over all age groups:
https://greatawakening.win/p/12iN7ImhT0/new-study-suggests-that-covid19-/
Not sure whether they subtracted out the flu deaths. They did remove the lockdown-related deaths (e.g., suicides, overdoses, etc.).
Further, there was the CDC report that ~96% of people of "COVID deaths" were people who had > 2.5 co-morbidities (i.e., they didn't die of COVID, just with it).
The effectiveness of Ivermectin/Fluvoxamine/HCQ/etc. (see others here) is 70-90% or so, when taken early.
So, in a sane world, where COVID treatments were easily available, the true mortality rate would be:
0.15% * 4% (removing co-morbities) * 30% (using a conservative estimate for those who would die even with one treatment) = 0.0018%.
That may or may not account for the removal of flu deaths.
That doesn't take into account that in a sane world, the doctor would use as many treatments as needed to save the patient (Dr. Flemming has done this for instance, and in a trial he only lost one patient out of dozens or hundreds, by using a mix-bag of treatments).
And that does not take into account the fact that in a sane world we would also be promoting vitamin D for nearly everyone.