Answer:
0.2 or 20%
Step-by-step explanation:
If the times of arrival vary uniformly, there is an equal chance of an employee reporting at any given time between 8:40 and 9:30.
The range between 8:40 and 9:30 is 50 minutes.
The range between 9:00 and 9:10 is 10 minutes.
Therefore, the probability that a randomly chosen employee reports to work between 9:00 and 9:10 is:
The probability is 0.2 or 20%.
13/15 - 1/3 Find a common denominator
1/3 x 5= 5/15
13/15 - 5/15 = 8/15
The answer is 4
how to show the work on here or how to send a picture of my notes :)
The decimal equivalent of 3/10 is .3
Answer: 0.5467
Step-by-step explanation:
Let X be the random variable that represents the income (in dollars) of a randomly selected person.
Given : 
Sample size : n=350
z-score : 
To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be
For x=15429
For x=15473
The required probability :-
Hence, the required probability is 0.5467.
