Q. Write the equation of the line.
1 answer:
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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)
- N is the sample size (6)
Thus, ME= ≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
Answer:
root(250)
answer choice C
Step-by-step explanation:
explanation in the pic above.
The population of the town from 2000 to 2020 is modeled by the equation:
Where t is the number of years since 2000.
We have to find the expected population of the town in 2002. The number of years from 2000 to 2002 are 2, so we have to find p(2). Replacing t by 2 in the above equation, we get:
So, according to the model, the population of town in 2002 is predicted to be 11,025.
Where t is the number of years since 2000.
We have to find the expected population of the town in 2002. The number of years from 2000 to 2002 are 2, so we have to find p(2). Replacing t by 2 in the above equation, we get:
So, according to the model, the population of town in 2002 is predicted to be 11,025.