Answer:
yes
Step-by-step explanation:
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
Im not sure if this is right but:
1. b
2. c
3. b
4. a
5. c
Answer:
c, vertical
Step-by-step explanation:
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