Answer:
Step-by-step explanation:
Raising e to both sides cancels out the natural logarithm on the left side and we have our exponential form. This works with other bases (recall that the natural logarithm, ln, has base e)
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
That would be the ones place, because 7 would be the tens place, 6 would be the hundreds place, and 5 would be the thousands place.
Hope this helped >.<
Just put it depends if they’re in a rush anything is possible
