EF = 6
EG = 21
You would subtract EF from EG to find FG:
21 - 6 = 15.
The answer is B.) 15
B is as it is a logical form of reasoning. if a=50, b=30, c=20
if a > b and b > c then a > c
Answer: (1.5, 3.5)
Explanation:
Find the slope by using the coordinates for points L and K.
3 - 4 = -1
5 - (-2) = 7
Slope = -1/7
Then you plot your points on the graph.
But you want to half your slope to find the midpoint. So your slope would now be -0.5/3.5.
And when you follow that slope, you will plot down the coordinate (1.5, 3.5), which is the midpoint.
Hope this helped! :3
(Also don’t mind the 19 at the top left)
Answer:
The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.
Step-by-step explanation:
The confidence interval for population variance is given as below:
![[(n - 1)\times S^{2} / X^{2} \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2} / X^{2} 1- \alpha/2, n- 1 ]](https://tex.z-dn.net/?f=%5B%28n%20-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20%20X%5E%7B2%7D%20%20%5Calpha%2F2%2C%20n-1%20%5D%20%3C%20%5Calpha%20%3C%20%5B%28n-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20X%5E%7B2%7D%201-%20%5Calpha%2F2%2C%20n-%201%20%5D)
We are given
Confidence level = 98%
Sample size = n = 81
Degrees of freedom = n – 1 = 80
Sample Variance = S^2 = 3.23
![X^{2}_{[\alpha/2, n - 1]} = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401](https://tex.z-dn.net/?f=X%5E%7B2%7D_%7B%5B%5Calpha%2F2%2C%20n%20-%201%5D%7D%20%20%20%3D%20112.3288%5C%5C%5CX%5E%7B2%7D%20_%7B1%20-%5Calpha%2F2%2Cn-%201%7D%20%3D%2053.5401)
(By using chi-square table)
[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]
[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]
2.3004 < σ^2 < 4.8263
Lower limit = 2.30
Upper limit = 4.83.
Answer:
no
Step-by-step explanation:
will take approximately 1-2 hours or even longer