If you are choosing class intervals in preparation for making a histogram, and if you know that the data all fall in the range from zero through 60 then you could choose the intervals:
0 through 10 inclusive
more than 10 to 20 inclusive
move than 20 to 30 inclusive
...
more than 50 to 60 inclusive
The classes listed have a class width of 11.
First, you decide how many intervals you need. (Can be between 5-15). Then, you find the range (Max-Min=Range) Next you divide the range by the amount of intervals you wanted. The quotient is the width of your data Example: Intervals(Width=10 and Minimum=28)(When you add the width to the minimum, you should count the minimum too. Instead of 28+10=38, it would be 37 since the 28 count, and to check the answer you have to add ten going down) 28-37 38-47 48-57 58-67 68-77
= If a rectangle is 10 meters longer than its width what is its width? =
The intervals must be of EQUAL SIZE.That is, if I have a data set of numbers ranging from 1 to 20, I might create 4 intervals of 5 each, because 20 divides by 5 evenly. So I would get a table:---------------------------------------------| Interval | Tally | Frequency |---------------------------------------------| 1-5 | | || | | || 6-10 | | || | | || 10-15 | | || | | || 15-20 | | |---------------------------------------------
15cm in width and 10 cm in length
The width of class intervals is normally determined by the data and not whether you want to calculate the arithmetic mean.For general population statistics you might use class widths of 6 or 10 years.But if you were studying school children in the UK you might use the following, unequal bands:Nursery-reception: 2 yearsYears 1 and 2: 2 yearsYears 3 to 6: 4 yearsYears 7 to 9: 3 years (Lower secondary)Years 10 to 11: 2 years (Upper secondary)Years 12 to 13: 2 years (A levels).
The classes listed have a class width of 11.
First, you decide how many intervals you need. (Can be between 5-15). Then, you find the range (Max-Min=Range) Next you divide the range by the amount of intervals you wanted. The quotient is the width of your data Example: Intervals(Width=10 and Minimum=28)(When you add the width to the minimum, you should count the minimum too. Instead of 28+10=38, it would be 37 since the 28 count, and to check the answer you have to add ten going down) 28-37 38-47 48-57 58-67 68-77
There is no requirement in the question that the classes should be of equal width. Consequently, class widths of 10, 10, 10, 10, 10 and 27 (or 30) will do the trick.
Make sure intervals start low enough to include the minimum value and extend high enough to include the maximum value. The intervals should not overlap or have any gaps. Between 5 and 10 intervals is reasonable.Greater than 10 or fewer than 5 intervals can result in a display that is not very helpful for seeing patterns in the data.
Intervals - album - was created on 2008-10-21.
= If a rectangle is 10 meters longer than its width what is its width? =
What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in
The weight comes in 10 pounds intervals with the highest at 200 pounds.
The width of a1969 Chevrolet C-10 is 79".
15cm in width and 10 cm in length
The intervals must be of EQUAL SIZE.That is, if I have a data set of numbers ranging from 1 to 20, I might create 4 intervals of 5 each, because 20 divides by 5 evenly. So I would get a table:---------------------------------------------| Interval | Tally | Frequency |---------------------------------------------| 1-5 | | || | | || 6-10 | | || | | || 10-15 | | || | | || 15-20 | | |---------------------------------------------