Answer:
I would but...
Step-by-step explanation:
where the question at
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%
Answer:
t^2 + 30t - 15/t^2 + 25
Step-by-step explanation:
85% of 40.
85/100 = .85
.85 * 40 = 34
Answer: 34
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,
