<span>Interval between 2.6 minutes late and 8.2 minutes late is within 2
standard deviations of the mean (k =8.2â’5.4/1.4 =5.4â’2.6/1.4 = 2). By Chebyshev’s Theorem, at least 1 â’1/2^2 =3/4 of all flights, i.e. at least 75% of all flights arrive
anywhere between 2.6 minutes late and 8.2 minutes late.</span>
Answer:
A. 
Step-by-step explanation:
Greatest Common Divisor [GCD]: 1½
I am joyous to assist you anytime.
Answer:
Step-by-step explanation:
1 km= 1000 m
1 dam= 10 m
8 km= 8000m
700 dam = 7000m
so Plane A
plane A is higher by 700m!
hope it helped!
Answer:
A 95% confidence interval for a population mean is computed from a sample of size 400. Another 95% confidence interval will be computed from a sample of size 100 drawn from the same population. Choose the best answer to fill in the blank: The interval from the sample of size 400 will be approximately _______as the interval from the sample of size 100.
Answer:
A is correct, x=7, AB=20
Step-by-step explanation:
All of the options say x=7 so you can just input 7 into 4x-8 to find AB
4(7)-8=AB
28-8=AB
20=AB
Hope this helps, please mark brainliest :)
