P = 2l + 2w
2w = p - 2l
w = (p - 2l)/2
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 answer should be 3.09%
Step-by-step explanation:371/120=3.09 and some change
9 loaves in each location with 8 loaves at one location
Answer:
x=5
Step-by-step explanation:
(1/81)^x=(243)^x-9
(1/3^4)=3^5(x-9)
3^-4x=3^5(x-9)
-4x=5(x-9)
-4x=5x-45
-4x-5x=-45
-9x=-45
x=5