Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.
We are to find the probability of selecting 1 red apple and 2 yellow apples.
Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.
Then, we have

Therefore, the probability of event A is given by

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.
Answer:
22.8
Step-by-step explanation:

Answer:
(x-2), (x+2), (3x-5)
Step-by-step explanation:
Factors of 3: ±1, ±3
Factors of 20: ±1, ±2, ±4, ±5, ±10, ±20
Possible factors of the polynomial: ±1, ±2, ±3, ±4, ±5, ±10, ±20, .... (there's a lot more but you probably do not need to list them all)
Pick a number to divide the polynomial by (I picked 2)
(3x³-5x²-12x+20)÷(x-2) = 3x²+x-10
So (x-2) is a factor of f(x) = 3x³-5x²-12x+20
Factor 3x²+x-10 = (3x-5)(x-2) these are the other factors of f(x) = 3x³-5x²-12x+20
3. 12 a
4. 7b-10
all you have to do is add up all of the sides.