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:
Step-by-step explanation:
Statements Reasons
AB // DC Given
AD // BC Given
AC ≅ AC Reflexive property
∠BAC ≅ ∠DCA Alternate interior angles
∠ACB ≅ ∠DAC Alternate interior angles
ΔCAB ≅ ΔDAC A S A
AD ≅ BC CPCT
Answer:b and c
Step-by-step explanation:
Bring the 3 to the other side, which makes 11-3 = 8
divide the whole equation by 2, which makes it Root (3x/5) = 8/2
square the whole equation which gives
, 3x/5 = 16
Multiply the equation by 5 which makes it, 3x = 80
finally divide the equation to find x, x= 80/3