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:
4:50 P.M.
Step-by-step explanation:
Sorry I don't know how to properly explain it but I just added 50 minutes to 4 o' clock to get the answer.
Hope this helps!
Maybe brainliest?
Answer:
The median, because the data distribution is skewed to the right
Step-by-step explanation:
If the longer part of the box is to the right (or above) the median, the data is said to be skewed right. If the longer part is to the left (or below) the median, the data is skewed left. The data is skewed right. The median would be a better estimate, because one or two numbers on the high end will cause the numbers to be skewed to the right, and the mean to be high
Answer:
113.04 cm³
Step-by-step explanation:
diameter = 6cm
so radius = 3cm
Volume of sphere = 

= 4 * 3.14 * 3 * 3 *3 / 3
= 4 * 28.26
= 113.04 cm³
Answer:
40
plss mark me brainliesttt
Step-by-step explanation:
4÷1/2=8
5×8=40