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
4*4-5*3-2x+6= -x-7/2 slope = 2
3 * 3 - 4 *2 -2x +1= - x - 2/2 slope = -2
2 + -2 = 0
Answer:
38 / 8 is 4.75
Step-by-step explanation:
Um I tried putting a photo of my work but it wouldn't let me sorry ;-;
I think it's t=778 not 100%
Answer:
3 is the quotient and 5 is the remainder
Step-by-step explanation: