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:
SR/BC=RT/CA And /R=/C S B And D
Step-by-step explanation:
Just Look at them then compare its pretty simple you just have to think about it.
8 people each buy 7 so, 8 x 7 = 56
She needs to sell 60:
60 - 56 = 4
Katy is 4 away from selling 60.
5n + (-6) = -2
add 6 to -2
5n = -2 + 6
5n = 4
divide both sides by 5
n = 4/5
1) 2 1/3 is also equal to 7/3
4 1/5 is also equal to 21/5
you need to make the denominators the same so you have to find the lowest common multiple, which is 15 since both 5 and 3 can be times by to make 15
you need to also times the numerators by the number you're timesing the denominator by
7/3 x 5 = 35/15
21/5 x 3 = 63/15
now add those together
35/15 + 63/15 = 98/15
now just convert 98/15 back to a mixed fraction
98/15 = 6 8/15
the answer is..
I'll do the others in a different answer!!
