Answered

A medical battery loses 5% of its power each week. How long will it take until only 20% of

its power remains.


Answer if you only know

Verified Answer :

\(\star\) \(20=100(0.95)^t\) 

\(\star\) \(0.2=0.95^t\) 

\(\star\) \(Log \ 0.2 = t \ log.95\) 

\(\star\) \(\dfrac{log0.2}{log.95} = t\) 

\(\star\) \(31.4 \ weeks = t\) 

\(\star\) \(Hence \ solved\)

Related Questions