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:
The Transcontinental Railroad
Step-by-step explanation:
Hope it helps you :)
Bill's Statement is Sometimes true.
Bill's statement is only true if 1 or both of the variables are positive but if both r and s are negative then it will be located on the left of r. Ex. r=-1 s=-2 (-1)+(-2)=-3
B subtract 18 from both sides
The constant of proportionality is 13
