A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees i s 12 inches with a standard deviation of 2.2 inches. Assume the distribution of diameters is roughly mound-shaped. What fraction of the trees will have diameters between 7.6 and 18.6 inches
1 answer:
Answer:
97.6% of the trees will have diameters between 7.6 and 18.6 inches.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12 inches
Standard Deviation, σ = 2.2 inches
We are given that the distribution of diameter of tree is a mound shaped distribution that is a normal distribution.
Formula:
P(diameters between 7.6 and 18.6 inches)
97.6% of the trees will have diameters between 7.6 and 18.6 inches.
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Answer:
exact Form:(0,),(1,3)
Equation Form: ( 0, )
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Step-by-step explanation:
3 : 5.....thats ur ratio...added = 8...so basically, Lisa wrapped 3 out of 8 (or 3/8) and Helen wrapped 5 out of 8 (or 5/8) 3/8(184) = 552/8 = 69 : Lisa wrapped 69 5/8(184) = 920/8 = 115 : Helen wrapped 115
Step-by-step explanation:
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Answer:
Option (b) is correct.
The expression is equivalent, but the term is not completely factored.
Step-by-step explanation:
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