The total number of tries = 10
The tries are { <span>110 111 100 000 101 111 100 000 011 010 }
</span><span />
Where: 0 representing heads and 1 representing tails
The tries which are heads came up more than once in 3 coin flips { 100 000 100 000 010 }
The number of the tries which are heads came up more than once in 3 coin flips = 5
∴ The probability of heads coming up more than once in 3 coin flips = 5/10 = 1/2
You have to figure out the highest common factor of each set of values:
9a + 21 = 3(3a + 7)
21b - 49 = 7(3b - 7)
54 - 6c = 6(9 - c)
8a + 32b = 8(a + 4b)
4p + 28q + 8r = 4(p + 7q +2r)
84a - 36b -12c = 6(14a - 6b - 2c)
6p + 9q + 15r = 3(2p + 3q + 5r)
18s - 30t + 54u = 6(3s - 5t + 9u)
Hope this helps :)
Answer:
v > -21
Step-by-step explanation:
-1/3v - 2 < 5
add 2
-1/3v < 7
multiply by -3
(if negative, switch sign)
v > -21
Answer:
This question cannot be answered with the information given
Step-by-step explanation:
You did not include the possibilities of each apple being selected.
I will follow this question and answer it when there's more details.
