Answer:
multiply by 10 on both sides
that is to get rid of the denominator
Answer: 0.6065
Step-by-step explanation:
Given : The machine's output is normally distributed with


Let x be the random variable that represents the output of machine .
z-score : 
For x= 21 ounces

For x= 28 ounces

Using the standard normal distribution table , we have
The p-value : 

Hence, the probability of filling a cup between 21 and 28 ounces= 0.6065
Rotation dilation translation reflection
Answer:
N/A
Step-by-step explanation:
This can't be answered, does not make sense, What percentage??