Those expressions are already in scientific notation.
Answer:
E1 and E2 are not mutually exclusive events. They have outcomes in common because a winning number can be both red and odd.
Step-by-step explanation:
EDMENTUM
Answer:
Ok, as i understand it:
for a point P = (x, y)
The values of x and y can be randomly chosen from the set {1, 2, ..., 10}
We want to find the probability that the point P lies on the second quadrant:
First, what type of points are located in the second quadrant?
We should have a value negative for x, and positive for y.
But in our set; {1, 2, ..., 10}, we have only positive values.
So x can not be negative, this means that the point can never be on the second quadrant.
So the probability is 0.
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
