If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then the probability of observing at least one car during any 5-minute time interval is 609/2500
Given The probability of observing at least one car on a highway during any 20 minute time interval is 609/625.
We have to find the probability of observing at least one car during any 5 minute time interval.
Probability is the likeliness of happening an event among all the events possible. It is calculated as number/ total number. Its value lies between 0 and 1.
Probability during 20 minutes interval=609/625
Probability during 1 minute interval=609/625*20
=609/12500
Probability during 5 minute interval=(609/12500)*5
=609/2500
Hence the probability of observing at least one car during any 5 minute time interval is 609/2500.
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The original number is 24
2 x 3 = 6
6 x 4 = 24
or
24 / 4 = 6
6 / 3 = 2
Answer:
The correct option is D. No, because a random sample from all customers of the shopping mall was not selected
Step-by-step explanation:
In statistics, Bias can be described as a term which depicts error if a sample is not taken evenly or it depicts errors taken from an unjust sampling.
In statistics, sampling bias means the errors which occur if one part of the population is favoured more then the rest of the populations. In this kind of bias, the individuals for experimentation are not chosen randomly.
As the customer satisfaction survey was distributed in only one of the gates hence, it does not give a generalized result and the result is biased.
Answer:
596.34m approx
Step-by-step explanation:
Given data
Let the Starting point be x
A helicopter flew north 325 meters from x
Then flew east 500 meters
Let us apply the Pythagoras theorem to solve for the resultant which is the distance from the starting position
x^2= 325^2+500^2
x^2=105625+250000
x^2= 355625
x= √355625
x=596.34m
Hence the distance from the starting point is 596.34m approx
Step-by-step explanation:
stating the ratios as fractions, setting the two fractions equal to each other, cross-multiplying, and solving the resulting equation
