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torisob [31]
3 years ago
7

What is 10 pie × 3 9 cow to the power of 10 10ths

Mathematics
1 answer:
spayn [35]3 years ago
7 0
The answer to this question is 21
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If one three-digit number ( 0 cannot be a left digit) is chosen at random from all those that can be made from the following set
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The probability of choosing a number that is not a multiple of 2 is P = 0.44

<h3 /><h3>How to find the probability?</h3>

We need to count the number of options for each digit.

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  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}.

The total number of combinations is the product between the numbers of options:

C = 8*9*9 = 648

If we want our number to not be a multiple of 2 then it must end in a odd digit, the combinations that meet that condition are:

  • For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
  • For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
  • For the third digit, we have 4 options {1, 3,  5, 7}.

C = 8*9*4 = 288

Then the probability of selecting a 3 digit number that is not a multiple of 2 is:

P = 288/648 = 0.44

If you want to learn more about probability, you can read:

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