Okay so this friend has a 8, 9, 10 as her options that she pays at.
8: 2,6 9: 6,3 10: 5,5
6,2 3,6 5,5
4,4
4,4
These are the possibilities of rolling these numbers. = 4 possibilities
4/11 (11 because that is the number of possibilities you get for two dice)
that leaves 7/11 possibilities of rolling and not paying!
36% lands on a spot that she pays at and about 64% possibilities of not paying.
Hope this helped!
:)
I’m sorry, I don’t know perpendicular but I would love to help.
Answer:
q=-5
Step-by-step explanation:
6+q=1
Isolate the variable by subtracting 6 from both sides of the equation
q=1-6
q=-5
Answer:
Therefore the required probability is 
Step-by-step explanation:
The probability of success is 
The number of trial = 4
X= the items survive out of 4
p =the probability of success and q = the probability failure.
p=
and 



Therefore the required probability is 
Step-by-step explanation:

#Hopeitshelp