First you have to make the assumption that these are the only two outcomes. There is also the possibility of hitting the ball and getting out.
However, if we assume that these are the only two cases, we know that the probability is 58.3%. This is because it has been on base 7 times out of 12.
Answer:
o.45
Step-by-step explanation:
any time you have a precent just put 0. infront of the number.
Answer:
11m + 4n
Step-by-step explanation:
when two numbers with the same variable are added together or subtracted the result is just the numbers added or subtracted. So you can add 6m and 5m to get 11m. and you can subtract 3n from 7n to get 4n
11m + 4n
Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.
Answer:
4(3m^3-2m^2+4m+2)
Step-by-step explanation:
