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
3x-10+x+8=90
4×-2=90
4x=92
X=23
One angle is 31. 23+8
The second one is 59. 3 (23)-10
Answer:
160
Step-by-step explanation:
72/45 = 1.6
1.6 * 45 = 72
Background information
160/100 = 1.6 (100 for percents)
160 fits the answer so it is right
HOPE THIS HELPS
PLZZ MARK BRAINLIEST
Answer:
10x+9
Step-by-step explanation:
You first have to multiply 3 into the parenthesis because you have to use distributive property.
Once you do that you get 3x+9+7x.
Combine like terms.
10x+9