I believe the answer is C
Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

For parts (a) and (b), you're given everything you need to determine

.
For part (c), if

and

are mutually exclusive, then

, so

. If the given probability is

, then you can find

. But if this given probability is for the intersection, finding

is impossible.
For part (d), if

and

are independent, then

.
Answer:
Your answer should be 1,200 because 120 x 10 is 1,200
<span>seven less than (-7) than the product of twice a number (2x) is greater than (>) five more than (+5) the same number (x)
2x-7>x+5
treat the > sign as an equals for now
2x-7=x+5
subtract x from both sides
x-7=5
add 7 to both sides
x=12
put the > sign back in
x>12</span>