to get the inverse of any expression, we start off by doing a quick switcheroo on the variables and then solve for "y".

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%
Step-by-step explanation:

Answer:
2/10
20/100
5/25
Step-by-step explanation:
I hope this helps you
If you would like to know what is y when x = 42, you can calculate this using the following steps:
x = 8 ... y = 20
x = 42 ... y = ?
8 * y = 20 * 42 /8
y = 20 * 42 / 8
y = 105
The correct result would be: y = 105 when x = 42.