Remember, the volume formula of a sphere is
.
Set that equation equal to whatever volume they give you and solve for r.
=
Now solve for r! I know you can do this. If you need help, just comment.
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:
8763
Step-by-step explanation:
Let x represent the number of students the college had last year. Then this year's enrollment is ...
... x - 3%·x = 8500
... x(1 - 0.03) = 8500 . . . . . collect terms
... x = 8500/0.97 ≈ 8762.89 . . . . divide by the coefficient of x
Enrollment last year was about 8763.
_____
Of course, you know 3% = 3/100 = 0.03.
Y + 0.4 = 2 .....subtract 0.4 from both sides
y = 2 - 0.4
y = 1.6 <==
Answer:
21.2
Step-by-step explanation:
