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:The area of the walkway around is 26 x + 16 as the sum of two terms
Step-by-step explanation:
Clearly, alternative B
y = 0.01x + 7.5
y = (0.01)*5000 + 7,5
y = 50 + 7.5
y = 57.50
Answer:
y≤-x+4
Step-by-step explanation:
x intercept =4
y intercept=4
the shaded area under the solid line because it is greater than or equal
A decimal is "." this symbol is used to separate whole numbers from a number value that is less than 1. Decimals basically puts things in its correct "place" according to the place value chart. | Example : $ 1.70 ; Without the decimal point this would be 170$ and in reality it is only one dollar and seventy cents. |