Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
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 (
)
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME=
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)
Thus, ME=
≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
2, No solution
3, Infinite many solutions
4, (0,-3)
i hope i was able to help!! so sorry if i made a mistake
Answer:
$15.71
Step-by-step explanation:
They have already raised $218.25.
$483.62 - $218.25 = $265.37 left to raise
There are 17 students, so the remaining amount can be divided between them.
$265.37 ÷ 17 students = $15.71 that each student needs to raise
You can check this by multiplying $15.71 by 17.
Answer:
24ft
Step-by-step explanation:
Pythagorean Theorem!!

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In this case, the ladder is c, and the base is b.
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Plug the numbers in the formula.

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Solve.





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