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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
Answer:
The probability that you would choose lemon-lime and then orange is 3/11 =.273.
Step-by-step explanation:
These are 'dependent events', which mean that your the event is affected by previous events. So, because you have eleven total bottles (five lemon-lime and six orange) and you do not replace the first bottle, that would only leave you with ten bottles remaining. The probability that you will pick the lemon-lime on the first choice is 5/11 because all of the bottles are there. However, your second choice will only include ten total bottles since you already took one. The probability that you would choose orange would be 6/10. When you multiply these two fractions and reduce to simplest form, you get 3/11.
Answer:
y=45
x=90
Step-by-step explanation:
x is a right angle, y is vertical from 45 degrees making it congruent to it.