=(-24÷6)×(-2)+7
=(-3×-2)+7
=6+7
=13
6.33985000288
i used a calculator
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<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:
(96.91, 101.09)
Step-by-step explanation:
To calculate the interval, you require the Z value at 95% confidence interval. The Z value is 1.96. The formula to calculate the confidence interval:
Where X is the mean and s is the standard deviation and x the sample size:
The interval is (96.91, 101.09)
