The number of ways it is possible for the gold, silver, and bronze medals to be awarded is 1320 ways
<h3>Permutation and Combination</h3>
Permutation are related to arrangement and combination has to do with selection.
If there are 12 teams, each representing a different country and the possible awards are gold, silver, and bronze medals, the number of ways they can be selected is given as:
12P3 = 12!/(12-3)!
12P3 = 12!/9!
12P3 = 12 * 11 * 10
12P3 = 1320 ways
Hence the number of ways it is possible for the gold, silver, and bronze medals to be awarded is 1320 ways
Learn more on permutation here: brainly.com/question/1216161
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1) 6÷0.2 = 30
If 6/2=3 then 6/0.2=30 as the decimal place shifts one place.
2)8÷0.1 = 80
8/1=8 so shift the decimal place over once to make 80.
3)9÷0.3 = 30
9/3=3 so shift the decimal place over once to get 30.
4)4÷0.04 = 100
4/4=1 so shift the decimal place over twice to get 100.
5)7÷0.002 = 3500
7/2=3.5 so shift the decimal place over three times to get 3500
6)0.718÷0.2 = 3.59
718/2=359 so shift the decimal over three places for the 0.718 and then back over once for the 0.2
7)0.0141÷0.003 = 4.7
141/3=47 so shift the decimal over our times for the 0.0141 and then back over three times for the 0.003
8)0.24÷0.012 = 20
24/12=2 so shift the decimal point over once twice for 0.24 then back over three times for 0.012
9)1.625÷0.0013 = 1250
1625/13=125 so shift the decimal point over three times for the 1.625 and then back four times for the 0.0013
10)47.1÷0.15 = 314
471/15=31.4 so shift the decimal point over once for the 47.1 and then back over twice for the 0.15.
Hope this helps :)
I think Theodosius the best measure. The outliers will not effect the measure. If you use mean, you add all the data points and divide by the number of data points. The extremes will skew the data. If you use mode, that may work also. You find the data that occurs most often.
D is the one that is decreasing
