Answer:
It equals -941.12.... There is a thing caled a calculator you can use for that ;D
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:
- P(x < 84) = 0.3085 or approximately 31%
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<em>Hello to you from Brainly team!</em>
<h3>Given</h3>
- Mean grade μ = 86,
- Standard deviation σ = 4,
- Grade limit x = 84.
<h3>To find </h3>
- Probability of that a randomly selected grade is less than 84 or P(x < 84).
<h3>Solution</h3>
Find z-score using relevant equation:
Substitute values and calculate:
Using the z-score table find the corresponding P- value.
- P(x < 84) = 0.30854 or approximately 31%
Answer:It's an arithmetic sequence with a common difference, d, = 4
A1 = 8
A2 = 8+4 = 12
A3 = 12+4 = 16 = 8+4(3-1) = 8+4(2) = 8+8
AN = A1 + d(N-1)
AN = 8 +4(N-1) = the Nth term in the sequence
Step-by-step explanation: hoped it helped