Answer:
its A!
Step-by-step explanation:
Answer:
2 3/4 C
Step-by-step explanation:

The answer is $3.57 because you must multiply 2.55 by 140 then divide that by 100 to find your answer hope this helps.
Answer:
31.82% probability that this day would be a winter day
Step-by-step explanation:
We use the conditional probability formula to solve this question. It is

In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening
In this question:
Event A: Rain
Event B: Winter day
Probability of rain:
0.42 of 0.25(winter), 0.23 of 0.25(spring), 0.16 of 0.25(summer) or 0.51 of 0.25(fall).
So

Intersection:
Rain on a winter day, which is 0.42 of 0.25. So

If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?

31.82% probability that this day would be a winter day