Y= DC-5, solve for D
1 answer:
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We are to find the probability that the weight of total luggage for a sample of 100 passengers is less than 2100.
The mean weight of the luggage of passengers will be 2100/100 = 21.
So we have to find the probability of the mean weight to be less than 21.
Average weight = u = 19.4
Standard deviation = 5.3
Since we are dealing with a sample of 100. We will use the standard error.
Standard error =
Now we have to convert the weight to z-score
From z table we can find the probability of z being less than 3.018 is 99.87%.
Therefore, the probability that for (a random sample of) 100 passengers, the total luggage weight is less than 2,100 lbs is 99.87%
The mean weight of the luggage of passengers will be 2100/100 = 21.
So we have to find the probability of the mean weight to be less than 21.
Average weight = u = 19.4
Standard deviation = 5.3
Since we are dealing with a sample of 100. We will use the standard error.
Standard error =
Now we have to convert the weight to z-score
From z table we can find the probability of z being less than 3.018 is 99.87%.
Therefore, the probability that for (a random sample of) 100 passengers, the total luggage weight is less than 2,100 lbs is 99.87%
Answer:
Mid - point = ( 2 , 1 )
Step-by-step explanation:
Let A = ( - 4 , 2 ) and B = ( 8 , 0 )
Let P = ( x , y ) be the mid-point of AB
Then,