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:
68% Confidence interval = [4.5752, 4.5848]
95% Confidence interval = [4.5688, 4.5918]
Step-by-step explanation:
Sample mean (X) = 4.580
Sample Standard Deviation (S) = 0.01065
Sample size (n) = 6
for alpha/2 0.84 = 1.1037
for alpha/2 0.975 = 2.5706
68% Confidence interval =
= [4.5752, 4.5848]
95% Confidence interval =
= [4.5688, 4.5918]
Answer: A, D
Step-by-step explanation:
68 is the initial amount.
A is correct because 0.2(68) represents 20% of 68. Since you're looking for 80% of 68, you subtract 0.2(68) from 68.
D is correct because it directly takes 80% of 68, which is what the question is asking for.
√30 = 5.47722557505
so 5.5ft × 5.5ft ≈ 30ft²
I hope I helped
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Answer:
I am not getting what is this