Answer:
[tex]t_{total}=985 h[/tex]
Explanation:
The time of use of the battery can de determined by:
[tex]t=0.98^n*20h[/tex]
t= is the time of duration
n=is the number of recharges
Now, we need to calculate n for a t=0h
[tex]t=0.98^n*20h[/tex]
[tex]\frac{t}{20h}=0.98^n[/tex]
[tex]ln(\frac{t}{20h})=ln(0.98^n)[/tex]
[tex]ln(\frac{t}{20h})=n*ln(0.98)[/tex]
[tex]n=ln\frac{ln(t/20h)}{ln(0.98)}[/tex]
We can apply the limit of t trending to 0 or we can stablish a value of t which we consider that the battery will be useless. For example t=0.1 h
[tex]n=ln\frac{ln(0.1h/20h)}{ln(0.98)}[/tex]
[tex]n=262[/tex]
For the total time:
[tex]t_{total}= \int_{0}^{262} 20h *0.98^{n}*dn[/tex]
[tex]t_{total}= 20h*(\frac{*0.98^{262}}{ln(0.98)}-\frac{*0.98^{0}}{ln(0.98)})[/tex]
[tex]t_{total}=985 h[/tex]