Answer: 6x
Work Shown:
For each step, the logs are all base b. This is to save time and hassle of writing tricky notation of having to write the smaller subscript 'b' multiple times. The first rule to use is that log(x^y) = y*log(x) for any base of a logarithm. The second rule is that meaning that the log base of itself is 1
log(b^(6x)) = 6x*log(b) .... pull down exponent using the first rule above
log(b^(6x)) = 6x*1 .... use the second rule mentioned
log(b^(6x)) = 6x
The mean and (sample) standard deviation σ = 0.2098.
<h3>What exactly would the standard deviation indicate?</h3>The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.
<h3>According to given information:</h3>The mean is the product of the dataset's total and the sample size. Mathematically.
The individual periods are Xi.
The sample size is N.
= 1.06 + 1.31 + 1.28 + 0.99,+ 1.48 + 1.37+ 0.98 + 1.31 + 1.59 + 1.55
= 12.92
N = 10
While substituting the value we get:
x = 12.96/10
x = 1.292
The samples' average is 1.292.
The standard deviation:
= (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.
= 0.43996
Putting into the formula we get:
σ = √(0.043996)
σ = 0.2098
The mean and (sample) standard deviation σ = 0.2098.
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I understand that the question you are looking for is:
You measure the period of a mass oscillating on a vertical spring ten times as follows:
Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55
Required:
What are the mean and (sample) standard deviation?
a. Mean: 1.228, Standard Deviation: 0.2135
b. Mean: 1.325, Standard Deviation: 0.1674
c. Mean: 1.292. Standard Deviation: 0.2211
d. Mean: 1.228, Standard Deviation: 0.2098
e. Mean: 1.292, Standard Deviation: 0.2135