By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)
<h3>How many more distance does the average race car travels than the average consumer car?</h3>
In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:
s' = (v' - v) · t
Where:
- v' - Speed of the average race car, in miles per hour.
- v - Speed of the average consumer car, in miles per hour.
- t - Time, in hours.
Please notice that a hour equal 3600 seconds. If we know that v' = 210 mi / h, v = 120 mi / h and t = 30 / 3600 h, then the distance surplus of the average race car is:
s' = (210 - 120) · (30 / 3600)
s' = 3 / 4 mi
The distance surplus of the average race car is equal to 3 / 4 miles.
To learn more on uniform rectilinear motion: brainly.com/question/10153269
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Answer:
Type I error.
Step-by-step explanation:
The decision to shut the process is triggered by the conclusion that the average height is significantly different from 66 mm.
This means that the null hypothesis, that states that the average height is not significantly different from 66 mm (μ=66), has been rejected.
If the null hypothesis is rejected, the error that can have been made is to reject a true null hypothesis, when the process is functioning to specification and the average length is not significantly different from 66.
This is a Type I error, that happens when a true null hypothesis is rejected.
Answer: 27.97 m^2
Explanation:
Area = Length x Width
Area = 3.6m x 7.77m
Area = 27.97 m^2
