<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:
If you could please write it down i will solve it for you.
Step-by-step explanation:
Answer:1.041 × 25 = 26.025 which rounds to $26.03 Hope this helped!
Step-by-step explanation:
Answer:
Step-by-step explanation:
(-6, 2) is the absolute minimum. There is no relative minimum.
It is 546.4583333333333, which would simplify to 546.46 if to the hundredths place or 546.5 to the tens place
