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:
9 + 2j is the required algebraic equation.
Answer:
y=3x-1/5
Step-by-step explanation:
we have to move 1/5 to the other side