Also, the order of the months is not important. For example, picking July and August is the same as picking August and July. So the combinations formula is used to solve this problem

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

There are 12 months, of which February, April, June, September and November, that is, 5 of them, have less than 31 months.

Desired outcomes:

Picking 2 months from a set of 5(those with less than 31 days). So

$D = C_{5,2} = \frac{5!}{2!(3)!} = 10$

Total outcomes:

Picking 2 months from a set of 12(all the months). So

$T = C_{12,2} = \frac{12!}{2!(10)!} = 66$

Probability:

$P = \frac{10}{66} = 0.1515$

15.15% probability that both months (different) have less than 31 days.

5 0

3 years ago

2 WILL MARK BRAINLIEST!!! HELP!!

elena55 [62]

Answer: A =19/2

Step-by-step explanation: Exact form a=19/2 decimal form a= 9.5 , mixed number a= 9 1/2

6 0

3 years ago

PLS HELP ITS DUE IN 5 MINUTES​

PLS HELP ITS DUE IN 5 MINUTES