He has 2/3 left to complete.
He did 5 out of 15 which is 5/15 or 1/3. Therefore he has 1-1/3 or 2/3 left.
SOLUTION
The mean is 4min
standard deviation is 1min
the z score is
where
then we have
The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
Answer:
<em>Hello bakugo here!</em>
Step-by-step explanation:
I don't see anything, is there supposed to be anything here? if not
<em>Thank you extra for FREE points! :D</em>