Answer:
The margin of error for the survey is 0.016
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 1024
Sample proportion:
We have to find the margin of error associated with a 90% Confidence interval.
Formula for margin of error:
Putting the values, we get:
Thus, the margin of error for the survey is 0.016
1 gal = 400 ft²
2 3/8 gal = ? ft²
to solve, use similar ratios with units gal/ft² and compare
(1)/(400) = (2 3/8)/(x)
*multiply both sides by (x) and divide both sides by (1)/(400), you get:
x = (400)*(2 3/8)/(1) = 400*(2 3/8) = 400*2 + 400*3/8 = 800 + 150 = 950 ft²
<u><em>answer is 950 ft²</em></u>
The rejection region is give by
where the test statistics is given by
i.e.
Thus,
Using the statistical table, the level of the test is 0.04.
Answer:
-2.92178
Step-by-step explanation:
Given the function 
The average,A is calculated using the formula;
![A=\frac{1}{b-a}\int\limits^a_b F(x)\, dx \\\\A=\frac{1}{7-1}\int\limits^7_1 3x \ Sin \ x\, dx \\\\\\=\frac{3}{6}\int\limits^7_1 x \ Sin \ x\, dx \\\\\#Integration\ by\ parts, u=x, v \prime=sin(x)\\=0.5[-xcos(x)-\int-cos(x)dx]\limits^7_1\\\\=0.5[-xcos(x)-(-sin(x))]\limits^7_1\\\\=0.5[-xcos(x)+sin(x)]\limits^7_1\\\\=0.5[-6.82595--0.98240]\\\\=-2.92178](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7Bb-a%7D%5Cint%5Climits%5Ea_b%20F%28x%29%5C%2C%20dx%20%5C%5C%5C%5CA%3D%5Cfrac%7B1%7D%7B7-1%7D%5Cint%5Climits%5E7_1%203x%20%5C%20Sin%20%5C%20x%5C%2C%20dx%20%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B3%7D%7B6%7D%5Cint%5Climits%5E7_1%20x%20%5C%20Sin%20%5C%20x%5C%2C%20dx%20%5C%5C%5C%5C%5C%23Integration%5C%20%20by%5C%20%20parts%2C%20u%3Dx%2C%20v%20%5Cprime%3Dsin%28x%29%5C%5C%3D0.5%5B-xcos%28x%29-%5Cint-cos%28x%29dx%5D%5Climits%5E7_1%5C%5C%5C%5C%3D0.5%5B-xcos%28x%29-%28-sin%28x%29%29%5D%5Climits%5E7_1%5C%5C%5C%5C%3D0.5%5B-xcos%28x%29%2Bsin%28x%29%5D%5Climits%5E7_1%5C%5C%5C%5C%3D0.5%5B-6.82595--0.98240%5D%5C%5C%5C%5C%3D-2.92178)
Hence, the average of the function is -2.92178
