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:
(3x + 10) / (2x + 5)(x - 5).
Step-by-step explanation:
(3/2x+5) + (5/x-5)
= [3(x - 5) + 5(2x + 5) ] / [ (2x + 5)(x - 5)]
= 3x - 15 + 10x + 25 / (2x + 5)(x - 5)
= 13x + 10 / (2x + 5)(x - 5).
Your answer is the first one q
Answer:
Step-by-step explanation:
Given
Total area of two plot 
Let l and b the length and width of plot
Perimeter for Fencing




Put value of b in 1




so there can be two value of l i.e. 
for