The steps to use to construct a frequency distribution table using sturge’s approximation is as below.
<h3>How to construct a frequency distribution table?</h3>
The steps to construct a frequency distribution table using Sturge's approximation are as follows;
Step 1: Find the range of the data: This is simply finding the difference between the largest and the smallest values.
Step 2; Take a decision on the approximate number of classes in which the given data are to be grouped. The formula for this is;
K = 1 + 3.322logN
where;
K= Number of classes
logN = Logarithm of the total number of observations.
Step 3; Determine the approximate class interval size: This is obtained by dividing the range of data by the number of classes and is denoted by h class interval size
Step 4; Locate the starting point: The lower class limit should take care of the smallest value in the raw data.
Step 5; Identify the remaining class boundaries: When you have gotten the lowest class boundary, then you can add the class interval size to the lower class boundary to get the upper class boundary.
Step 6; Distribute the data into respective classes:
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Answer:
V = 500 pi in^3
or approximately 1570 in ^3
Step-by-step explanation:
The volume of a cylinder is given by
V = pi r^2 h where r is the radius and h is the height
The diameter is 10. so the radius is d/2 = 10/2 =5
V = pi (5)^2 * 20
V = pi *25*20
V = 500 pi in^3
We can approximate pi by 3.14
V = 3.14 * 500
V = 1570 in ^3
Answer:
Step-by-step explanation:
Vertical lines take the form of 
and horizontal lines take the form of 
Why is this? Notice that whenever you make a horizontal line on a graph and select virtually any point that's on that line, all of the points you pick will have the same y value. Try to imagine the same thing with a vertical line, are the results the same?
So the line (we got -12 from the coordinate that was given to us on the question prompt) is a vertical line that passes through virtually all possible y values (from positive infinity to negative infinity, or vice-versa) which means that the question prompt having given us the y value was negligible.
Good luck!
F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
