Number of ways the first 4 databases can be selected = 9C4 = 126
Probability that it is in 2 of the first 4 = 5C2 / 126 = 10/126
Probability that it is in 3 of the first 4 = 5C3 / 126 = 10/126
Probability that it is in 4 of the first 4 = 5C4 / 126 = 5/126
Probability that it is in at least 2 of the first 4 = 10/126 + 10/126 + 5/126 = 25/126
A discontinuity occurs when the denominator equals zero
it will be either a hole or an asymptote
x+4 = 0
subtract 4 from each side
x=-4
Answer:
2/15
Step-by-step explanation:
4/5-2/3 = 12/15-10/15 = 2/15
So 2/15 is your answer
Answer:
167
Step-by-step explanation:
the change between 276 and 109 is 167