What is 999999999 times 999999999
2 answers:
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Answer:
<u>x = 8</u>
Step-by-step explanation:
The sides are in proportion.
- 40 / 7x + 14 = 44 / 33 + 44
- 40 / 7x + 14 = 44/77 = 4/7
- 40(7) = 4(7x + 14)
- 280 = 28x + 56
- 10 = x + 2
- <u>x = 8</u>
Answer:
a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.
b)
The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.
c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.
Question a:
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 100 - 1 = 99
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.9842
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.
The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.
The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.
Question b:
Critical value is the same(same sample size and confidence level).
The margin of error is:
The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.
The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.
The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.
c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).
They are respected, as the upper bound of both intervals is below the new FAA recommendations.