The answers should be 1, 1, 0, 0
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 answer is on photomath
Answer:
The answer to your question is: 9.01 x 10⁸ ice cream cones
Step-by-step explanation:
Data
# of ice cream cones = 1700
5.3 x 10 ⁵ minutes ---------- 1 year
# of ice cream cones purchased in one year = ?
1700 cones --------------------- 1 min
x ---------------------- 5.3 x 10 ⁵ minutes
x = (5.3 x 10 ⁵)(1700)/1 = 901 000 000 ice cream cones
In scientific notation
9.01 x 10⁸ ice cream cones
Multiply the values then either divide by 2 or multiply by 2