Answer: 130
Step-by-step explanation:
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:
(x-3)(x^2-7) = x^3 -3x^2 -7x + 21
just change the sign of the root and put x in front of it, then multiply the factors all together
(x-3)(x-sqr7)(x+sqr7) roots come in conjugate pairs to eliminate irrational coefficients
(x-sqr7)(x+sqr7) = x^2-7. similar to (a-b)(a+b) = a^2-b^2
Step-by-step explanation:
I think its 9000(1.045)^5. Hope this helps. That equals $11,215.60, or rounded to the nearest dollar, $11,215.
First, reorder the numbers from least to greatest
1.4, 1.6, 1.7, 2.5, 2.5
The median is the middle number
1.7 is your answer
hope this helps