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What is the probability that a two-digit number selected at random has a tens digit less than or equal to its units digit? Express your answer as a common fraction.

 Dec 25, 2020
 #1
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0

54 out of 90

28/45

 Dec 25, 2020
 #2
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27/45     or  3/5

 Dec 25, 2020
 #3
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11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 33 , 34 , 35 , 36 , 37 , 38 , 39 , 44 , 45 , 46 , 47 , 48 , 49 , 55 , 56 , 57 , 58 , 59 , 66 , 67 , 68 , 69 , 77 , 78 , 79 , 88 , 89 , 99 , Total =  45 such numbers.

 

Probability = 45 / 90 = 1 / 2

 Dec 25, 2020
 #4
avatar+114320 
+1

Note  the pattern :

 

99  = 1

88,89  = 2

77, 78, 79  = 3

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11, 12, 13, 14, 15, 16, 17, 18, 19  =  9

 

 

The number of  two digit numbers  will just be the sum of the  first nine positive integers   =

 

(9)(10)/ 2  =   45

 

And we have  (99 - 10 + 1)  =   90  two digit numbers

 

P =   45/90      =  1/2

 

cool cool cool

 Dec 25, 2020

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