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2 answers:
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Answer:
as much as I can go
Step-by-step explanation:
we have
we notice that we have as a common factor
so
Answer:
Option D. 938/969
Step-by-step explanation:
At most 2 defective means 2 or less than 2 bulbs are defective
So, we have 3 cases:
a) No defective bulb. b) 1 defective bulb. c) 2 defective bulbs
Case a) No defective bulb
Total number of bulbs = 20
Number of defective bulbs = 5
Number of non-defective bulbs = 15
Total number of ways to select 0 defective bulb = 15C4 = 1365
Case b) 1 defective bulb
Total number of ways to select 1 defective bulb = 15C3 x 5C1 = 2275
Case c) 2 defective bulbs
Total number of ways to select 2 defective bulbs = 15C2 x 5C2 = 1050
Therefore, total number of ways to select at most 2 defective bulbs = 1365 + 2275 + 1050 = 4690
Total number of ways to select 4 bulbs from 20 = 20C4 = 4845
Therefore, probability of selecting at most 2 defective bulbs =
Therefore, option D gives the correct answer.
Answer:
Your answer is C. 8.
Step-by-step explanation:
Amplitude: 2
Period: 8
Vertical Shift: 3
