Answer:
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Total number of ways to make a pair:
The first player can be any one of 7 . For each of those . . .
The opponent can be any one of the remaining 6 .
Total ways to make a pair = 7 x 6 = 42 ways .
BUT ... every pair can be made in two ways ... A vs B or B vs A .
So 42 'ways' make only (42/2) = 21 different pairs.
If every pair plays 2 matches, then (21 x 2) = <em><u>42 total matches</u></em> will be played.
Now, is that an elegant solution or what !
60 children tickets and 190 adult tickets were sold.
Step-by-step explanation:
Let the no. of adult tickets sold be 'a'
Let the no. of children tickets sold be 'c'
Total tickets sold = 250
Cost of 1 children ticket = $2.5
Cost of 1 adult ticket = $4
Total money collected= $910
Given that,
a + c = 250
a = 250 - c
4a + 2.5c = 910
Substitute a value
4(250 - c) + 2.5c = 910
1000 - 4c + 2.5c = 910
1000 - 1.5c = 910
-1. 5c = -90
1.5c = 90
c = 90/1.5
c = 60
a + c = 250
a + 60 = 250
a = 190
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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