What is the solution to the following equation?
2 answers:
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The length of the curve from x = 3 to x = 6 is 192 units
The curve is given as:
from x = 3 to x = 6
Start by differentiating the curve function
Evaluate
The length of the curve is calculated using:
This gives
Expand
This gives
Express as a perfect square
Evaluate the exponent
Differentiate
Expand
L = (6³ + 6) - (3³ + 3)
Evaluate
L = 192
Hence, the length of the curve is 192 units
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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