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.
- 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 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:
brainly.com/question/251701
Answer:
Step-by-step explanation:
- Well I hate to break the news but 243 is not a perfect square. I'll work you through it, 243 is not a perfect square because it is not an even number. an even number must end in (0,2,4,6,8)
- Step one. Find the square root. the square root of 243 is <em>15.588. </em>
- Step two Is it a perfect square. No 243 just cant be a perfect square.
- Hope this helped :)
If you multiply 3 and 3 and 3
The answer is on the paper
