Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where

M is the sample mean weight of the wild mountain lions ( $\frac{69 +104 +125 +129+60 +64}{6} =91.8$ )

ME is the margin of error of the mean

And margin of error (ME) of the mean can be calculated using the formula

ME= $\frac{t*s}{\sqrt{N} }$ where

t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)

s is the standard deviation of the sample(31.4)

N is the sample size (6)

Thus, ME= $\frac{1.30*31.4}{\sqrt{6} }$ ≈16.66

Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5

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3 years ago

Ashley bought bag of oranges at the grocery store she gave half of the oranges to David then she gave Erin 6 oranges and Kept th

andreev551 [17]

The answer would be 16. She gave Erin 6 and kept two. That would be 8 oranges between the two of them and David got half, making the total 16 oranges.

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3 years ago

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Expand or factor each of the following expressions to determine which expressions are equivalent.

bearhunter [10]

(3x-4y)^2=9x^2-24x+16y^2
(3x-2)(9x^2+6x+4)=27x^3-8

4 0

3 years ago

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