Answer:45 10-5-108, as last amended by Laws of Utah 2017, Chapter 193 ... 247 (d) Title 63G, Chapter 2, Government Records Access and Management Act; 248 (e)
Step-by-step explanation:
<u>Answer with explanation:</u>
As per given , we have
A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72.
i.e. n= 50
Degree of freedom = n-1=49


The point estimate of the population mean is equals to the sample mean.
i.e. The best estimate for the length of gribbles= 3.1 mm
Also, population standard deviation is unknown , then the confidence interval would be :

For df= 49 and significance level of 0.05 , the t- value (using t-distribution table) would be :

Now, Margin of error : 

95% confidence interval : 

Written in scientific notation that is:
<span>86,900.4
= 8.6x10^4</span>
Answer:
15, 840.
Step-by-step explanation:
-HCM(GCM)
105's factor are 1, 3, 5, 7, <em>15</em>, 21, 35, 105.
125's factor are 1, 2, 3, 4, 5, 6, 8, 10, 12, <em>15</em>, 20, 24, 30, 40, 60, 120.
so, the answer is 15!