A frozen food company uses a machine that packages okra in six ounce portions. A sample of 5454 packages of okra has a variance of 0.440.44. Construct the 98%98% confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two decimal places.

98% confidence interval: (0.29, 0.73)      

Step-by-step explanation:

We are given the following in the question:

Sample size,n = 54

Sample variance =

Significance level = 0.02

Degree of freedom = n - 1 = 53

We use chi-square to find the confidence interval for population variance.

Formula:

Putting values, we get,

is the required 98% confidence interval for the variance of the weights of the packages prepared by the machine.

okay wait before i answer, is this for your points on a graph?

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.