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Posts: 158
In response to
"
probably -- nm
"
by
znufrii
I'm thinking something along the lines of...
Posted by
znufrii
Apr 26 '13, 10:57
dA = (t-A)/n
where
dA = difference in average due to latest install
t = time of latest install
A = previous average
n = total number on installs
Responses:
(note: this may be entirely incorrect) -- nm
-
znufrii
Apr 26, 10:57
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