the answer to this question is 500 dimples on a golf ball.
The answer is: <span>t=17.5 and f=12
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Hope this helps!!
~Lena~
320 i believe... the original answer was 317.52
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
1) Distribute
2x + 28x + 21
2) Combine like terms
30x + 21
Since there are no more like terms, 30x and 21 cannot be combined. Therefore, the answer is 30x+ 21