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Answer:
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Step-by-step explanation:
We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.
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 = 35 - 1 = 35
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of
). So we have T = 2.0322
The margin of error is:
M = T*s = 2.0322*30 = 60.97
The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Answer:
BE = FC = 3 inches, EF = 2 inches
Step-by-step explanation:
The sum of angles A and D is 180°, so the sum of their half-angles is 90°. That is, half of A plus half of B add to 90°, so the bisector from B intersects AE at a right angle. Call that point of intersection X.
Then angle ABX = angle EBX, so triangle ABX is congruent to triangle EBX. Sides AB and BE are corresponding sides of congruent triangles.
The same argument applies to sides DC and CF.
Thus we have BE = CF = 3 inches, and EF is the left-over distance, 2 inches.
Answer:
The standard deviation of the age distribution is 6.2899 years.
Step-by-step explanation:
The formula to compute the standard deviation is:

The data provided is:
X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}
Compute the mean of the data as follows:

![=\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B11%7D%5Ctimes%20%5B19%2B19%2B21%2B...%2B40%5D%5C%5C%5C%5C%3D%5Cfrac%7B299%7D%7B11%7D%5C%5C%5C%5C%3D27.182)
Compute the standard deviation as follows:

![=\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B11-1%7D%5Ctimes%20%5B%2819-27.182%29%5E%7B2%7D%2B%2819-27.182%29%5E%7B2%7D%2B...%2B%2840-27.182%29%5E%7B2%7D%5D%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%5Cfrac%7B395.6364%7D%7B10%7D%7D%5C%5C%5C%5C%3D6.28996%5C%5C%5C%5C%5Capprox%206.2899)
Thus, the standard deviation of the age distribution is 6.2899 years.
Answer:
Step-by-step explanation:
Use the following formula:
slope: 
Let:

Plug in the corresponding numbers to the corresponding variables:

is your answer.
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