Cuanto tiempo tomara para que estén presentes 875 bacterias?
875=300(4/3)^t
875/300=(4/3)^t
log4/3(875/300) = t
log(875/300) / log(4/3) = t
t = ~3,7209180353
Aynnnnnnssssssss no c :p
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875=300(4/3)^t
875/300=(4/3)^t
log4/3(875/300) = t
log(875/300) / log(4/3) = t
t = ~3,7209180353
Aynnnnnnssssssss no c :p