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.
4 gigabytes because 71-50= 21 4*5=20
Answer:
x-int= (18,0), y-int= (0,-6)
Hello there!
First, let's categorize our monomials with the proper variables.
3x + 12x
11y - 5y
-10z -4z
Now, we can make these into simpler terms by adding/subtracting the sectioned variables.
3x + 12x = 15x
11y - 5y = 6y
-10z - 4z = -14z
Now let's place our results into an expression as a final answer.
15x + 6y - 14z is your final expression.
I hope this helps!
10850 would be I'm favor of building the park and 4650 would not