Answer:
The best point of estimate for the true mean is:

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.
Step-by-step explanation:
Information given
represent the sample mean for the late time for a flight
population mean
represent the population deviation
n=76 represent the sample size
Confidence interval
The best point of estimate for the true mean is:

The confidence interval for the true mean is given by:
(1)
The Confidence level given is 0.95 or 95%, th significance would be
and
. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got
Replacing we got:
Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.
Answer:
Step-by-step explanation:
- Speed in one direction = 40 km/h
- Speed on return = x
- Average speed = 60 km/h
Let the distance be d
<u>Average speed </u>
- 2d/(d/40 + d/x) = 60
- d/40 + d/x = 2d/60
- 1/40 + 1/x = 1/30
- 1/x = 1/30 - 1/40
- 1/x = (4-3)/120
- 1/x = 1/120
- x = 120 km/h
Answer:
- large: 18.5 kg
- small: 15.75 kg
Step-by-step explanation:
Let b and s represent the weights of the big and small boxes, respectively. Then the two delivered weights can be summarized as ...
5b +6s = 187
3b +2s = 87
We can eliminate the "s" variable by subtracting the first equation from 3 times the second:
3(3b +2s) -(5b +6s) = 3(87) -(187)
4b = 74 . . . . . collect terms
b = 18.5 . . . . . divide by 4
Using this value in the second equation, we find ...
3(18.5) +2s = 87
2s = 31.5 . . . . . . . . subtract 55.5
s = 15.75 . . . . . . . . divide by 2
The large box weighs 18.5 kg; the small box weighs 15.75 kg.
Answer:
24
Step-by-step explanation:
In a deck of 52 cards, there are 4 kings and 4 queens
Selection of 3 kings out of 4 = 4 C 3 = 4
Selection of 2 queens out of 4 = 6
So, total number of selections = 4 x 6 = 24