Answer:
Step-by-step explanation:
1.put your percent over a 100 so
125%/100 then put 60 over x
2.then cross multiply 100x60=6000
3.then divide 6000 by 125 which is percent
4.your answer is-48
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%
The answer for this is B because
GEF
The middle point aka the E has to be in the middle and then the first and last points don’t matter
Answer:
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