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
Use the formula a = a+b/2h
Answer:
rhombus
Step-by-step explanation:
Rhombus has that equiangular and is equilateral but I'm not entirely sure about this...
X^2 + 8x + 16 =
(x + 4)(x + 4) or (x + 4)^2
Let e=erasers and p=pencils. So we know the total amount of erasers plus pencils sold is equal to 220. Therefore,

We also know the cost of erasers and pencils totaled to $695 in earnings and so:

and so those are your two equations.