Ashley took $20 out on her savings account each week on Friday. How many weeks ago did she have at least $250 in her account
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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
In this question , we have to write an arithmetic expression that calculates the average of 18 and 46.
TO find the average, we have to add the numbers and divide by 2.
So here we have to add 18 and 46 and divide by 2, that is
And that's the required algebraic expression .