It’s B
PEMDAS. Exponents first then divide then add
36 equally-likely outcome: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1),(6,2), (6,3), (6,4), (6,5), (6,6)
Solution:
Outcomes with first number being old number and second being even number: (1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6) = 9 outcomes
P(old,even) = 9/36 =1/4 = 0.25
For the answer to the question above,
It seems the function f(x) is 3/(x+4).
If the function is f(x) = 3 / (x + 4), a table could be
x f(x)
-8 -3/4
-7 -1
-6 -3/2
-5 -1
(-4 is no a valid input)
- 3 3
- 2 3/2
-1 1
0 3/4
1 3/5
2 3/6 = 1/2
3 3/7
4 3/8
If the function is f(x) = (3/x) + 4
Just substitute different values for x and obtain the respective outputs. In this case, x has to be different of o, i.e. 0 is not a valid input.
