I think it would be 26. I hope that helped.
Really hope i'm right
1st row
the first one is length less than 10
one next to it is width greater than 5
next to that is length greater than 10
2nd row
the first one is length greater than 10
one next to that is width less than 5
next to that is width greater than 5
3rd row
first one is width less than 5
and last is length less than 10
Answer:
χ²R = 8.643
χ²L = 42.796
0.20 < σ < 0.45
Step-by-step explanation:
Given :
Sample size, n = 23
The degree of freedom, df = n - 1 = 23 - 1 = 22
At α - level = 99%
For χ²R ; 1 - (1 - 0.99)/2= 0.995 ; df = 22 ; χ²R = 8.643
For χ²L ; (1 - 0.99)/2 = 0.005 ; df = 22 ; χ²L = 42.796
The confidence interval of σ ;
s * √[(n-1)/χ²L] < σ < s * √[(n-1)/χ²R)]
0.28 * √(22/42.796) < σ < 0.28 * √(22/8.643)
0.2008 < σ < 0.4467
0.20 < σ < 0.45
Y=-5.11111 how I solved it is in the picture attached
(c - 3)(2c - 1) is the factored form
