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MAXImum [283]
2 years ago
15

The weight of oranges growing in an orchard is normally distributed with a mean weight of 6.5 oz. and a standard deviation of 1.

5 oz. Using the empirical rule, determine what interval would represent weights of the middle 99.7% of all oranges from this orchard.
Mathematics
1 answer:
sergey [27]2 years ago
6 0
Add 3 standard deviations above and below the mean to get the range in which 99.7% of the data in a normal distribution will fall
6.5 + 4.5 = 11
6.5 - 4.5 = 2
So 2 to 11 ounces would be the interval
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The answer is "Option C"

Step-by-step explanation:

please find the complete question in the attached file:

Its ratio of samples by Johns Hopkins will be about the equivalent than those from Ohio State because sample varying depending on the sample, each of them would have the same variability also like the amount, that's why he assumes the variance of sample sizing in the sample percentage p with both the hat above, relative to the confidence interval in Ohio State determined from its Johns Hopkins test.

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I dont know if this is right but I think its 125/343

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Ede4ka [16]

Answer:Cross multiply and solve for x.

x/6 = 5/1.25

let the height of building be h ft

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Step-by-step explanation:

Well first let’s covert the ft to in- 5ft is 60 in and 6ft is 72 in. Then we want to set up 2 fractions based on height and shadow, so we have 60/15=x/72 (we put x because that’s what we’re trying to find). Then we do cross multiplication so now we have 15x=4320. We divide 15, so now we have 4320/15=288. So the building is 288 in tall or 24ft tall! Glad to help! :)

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