8; you subtract the largest number (in this case 9) with the smallest number (1) and
9-1= 8
Subtract the smallest number in the data set from the largest number in the data set. eg, the range of {3, 1, 5, 6, 6 ,8, 10, 5, 8} is 10 - 1 = 9
If the set of numbers is sorted: 1 1 2 3 4 5 5, then you can see that 1 and 5 both lie at the outer limits of the range of data.
the range for 5 2 1 and 3 is 4.
Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.
find the range 3 -9 7 -1 5 -4 2
To find the range of a data set, you first need to order the data from smallest to largest. In this case, the data is already ordered in such a way. The next step is to take the least (-4) and greatest (1) numbers, and subtract the least from the greatest. In this case, we'd do 1-(-4) which equals 5. Thus the range of the data set in the question is 5.
The range is 6. (6 - 0 = 6)
Subtract the smallest number in the data set from the largest number in the data set. eg, the range of {3, 1, 5, 6, 6 ,8, 10, 5, 8} is 10 - 1 = 9
The number 5.
8
The range is 17: 20 - 3 = 17
seven
Yes, but the two are measures of very different things. The median is a measure of central tendency whereas the range is a measure of spread. Nevertheless, the set 1, 2, 3, 4, 4 has a range of 3 and a median of 3.
The meaning to range is minusing the largest number by the smallest number. eg (1 1 1 2 2 3 3 3 4 5 5 6 6 7 7 8 9 9) 9-1 =8
The range of database application are given below: 1- Personal Database 2- Workgroup Database 3- Department Database 4- Enterprise Database
Factors of 36:36 = 22 * 32Thus, n and 36 has highest common factor as 1, if and only if when n is not a multiple of 2 and 3.Total number of multiples of 2 in the given range=500Total number of multiples of 3 in the given range=333Total number of multiples of 6 in the given range=166Total number of multiples of 2 and 3 in the given range=500+333-166=667Thus, the total number of 'n' in the given range which are not multiples of 2 and 3= 999-667 = 332.332 numbers are there such that it is greater than or equal to 2 and less than or equal to 1000 and the highest common factor of the number and 36 is 1.
I think you mean ordinal data. Similar to the golf tournament, you need to determine where to "cut" (from the ordinal data) so as to divide the data into different categories (to the nominal data). For example, if the ordinal data range from 1 to 6 (where 1 = the best) and the cut is 3, then you convert all the numbers from 1 to 3 to "1" (which represents "good") and the all numbers from 4 to 6 to "2" (which represents "bad"). In other words, 1, 2, and 3 from the original ordinal data set are converted to "1" (ordinal data); whereas 4, 5, and 6 from the original date set now become "2" (ordinal data). Eddie T.C. Lam