The formula for the volume of a cube is V(s) = s3 where s is the side length of the cube. In which quadrant(s) is the graph of t
2 answers:
Answer:
The answer is the option A
quadrant I
Step-by-step explanation:
Let
s-----> the length side of a cube
V----> the volume of a cube
we know that
the volume of a cube is equal to
The volume in function notation is
Remember that
The volume and the side length can not be a negative number
The domain and the range of the function are all real positive numbers
so
the graph of the function V(s) belong to the I quadrant
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Answer: Probability that the proportion of students who graduated is greater than 0.743 is P = 0.4755
Step-by-step explanation:
Given that,
Probability of freshmen entering public high schools in 2006 graduated with their class in 2010, p = 0.74
Random sample of freshman, n = 81
Utilizing central limit theorem,
So,
= P( Z > 0.0616)
= 0.4755 ⇒ probability that the proportion of students who graduated is greater than 0.743.
Answer:
1) Probability that all five are good = 0.46
2) P(at most 2 defective) = 0.99
3) Pr(at least 1 defective) = 0.54
Step-by-step explanation:
The total number of bulbs = 30
Number of defective bulbs = 4
Number of good bulbs = 30 - 4 = 26
Number of ways of selecting 5 bulbs from 30 bulbs =
ways
Number of ways of selecting 5 good bulbs from 26 bulbs =
ways
Probability that all five are good = 65780/142506
Probability that all five are good = 0.46
2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)
Pr(no defective) has been calculated above = 0.46
Pr(1 defective) =
Pr(1 defective) = (14950*4)/142506
Pr(1 defective) =0.42
Pr(2 defective) =
Pr(2 defective) = (2600 *6)/142506
Pr(2 defective) = 0.11
P(at most 2 defective) = 0.46 + 0.42 + 0.11
P(at most 2 defective) = 0.99
3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)
Pr(1 defective) =0.42
Pr(2 defective) = 0.11
Pr(3 defective) =
Pr(3 defective) = 0.009
Pr(4 defective) =
Pr(4 defective) = 0.00018
Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018
Pr(at least 1 defective) = 0.54
Answer:
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Step-by-step explanation: