Answer : 0.0129
Step-by-step explanation:
Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 
years and standard deviation is 
years.
Sample size : 
Let X be the random variable that represents the age of fleets.
We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.
For z-score,
For x=14
By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-
Hence, the required probability = 0.0129
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Answer:
1800 miles
Step-by-step explanation:
No. of miles driven by Mr. Thomas in May = 75
It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).
Thus
No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May = 6*75 = 450 miles.
__________________________________________________
Another condition given that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).
Thus
No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July = 4* 450 miles = 1800 miles.
No. of miles driven by Mr. Thomas in June is 1800 miles.
Step-by-step explanation:
A
=
15
I hope this is correct
10 more than (+) twice a number (2x) is (=) 36
10 + 2x = 36 Subtract 10 from both sides
2x = 26 Divide both sides by 2
x = 13
