If You Have Numbers: 2,2,2,5,5,7,7,8,11 You Add All Of The Numbers: 2+2+2+5+5+7+7+8+11=49 Then You Divide (Hint: Count How Much Numbers You Have In The Set Of Numbers) Then Divide By How Much Numbers You Counted: 49 Divided By 9 = 5.4444444 And So On So The Meadian Is 5.4444444 And So On That Is How You Get The Meadian In A Set Of Numbers* * * * * The above answer is totally incorrect. It gives the method for finding the MEAN which is not the MEDIAN. To find the median, put the numbers in ascending order and then find the middle one. If there are an even number of values, find the average (mean) of the middle two. So for the above example, there are 9 numbers so the middle one is the 5th. So the median is the 5th number = 5. f there was another 11, there would be 10 numbers and the median would be the average of the 5th and 6th numbers = (5+7)/2 = 6
f = 54
42 losers
25+11=36: Let f and s represent the first and second numbers respectively. The statement of the problem yields two equations: f + s =36 and f = 3 + 2s. Substituting the function given in the second equation for f into the first equation yields 3 + 2s + s = 36, or (subtracting 3 from each side and merging the s terms, 3s = 33 or s = 11. Then f + 11 = 36 (substituting the value for s into the first equation), or f = 25.
f = {n : n Î Z, n ≥ 12}
If You Have Numbers: 2,2,2,5,5,7,7,8,11 You Add All Of The Numbers: 2+2+2+5+5+7+7+8+11=49 Then You Divide (Hint: Count How Much Numbers You Have In The Set Of Numbers) Then Divide By How Much Numbers You Counted: 49 Divided By 9 = 5.4444444 And So On So The Meadian Is 5.4444444 And So On That Is How You Get The Meadian In A Set Of Numbers* * * * * The above answer is totally incorrect. It gives the method for finding the MEAN which is not the MEDIAN. To find the median, put the numbers in ascending order and then find the middle one. If there are an even number of values, find the average (mean) of the middle two. So for the above example, there are 9 numbers so the middle one is the 5th. So the median is the 5th number = 5. f there was another 11, there would be 10 numbers and the median would be the average of the 5th and 6th numbers = (5+7)/2 = 6
(f-1)'=2+2x,x is greater than or equal to zero and f'(2)=3,find f'(x)
There is no simple answer. Suppose the numbers, in non-decreasing order are A, B, C, D, E and F then A = B = 21 25 < C < 30 D = 60 - C D < E < 35 and F = 35. There are, thus, infinitely many possible values for C, and hence for the data set.
No. The set of x-values are the domain for only g. This will result in a set of images, which will be g(x). This set of g(x) values are the domain of f.
False
The formula for calculating median on grouped data is L + I *(50% * N-F)/f L - lower limit of the median group I - group interval N - Number of frequency (total sum of frequencies in each group) F - cumulative freqency for the group before the median group f - frequency of the median group Since median is just the same as the 2nd quartile, we use 0.5 in place of 50% in the formula. We can tweak the formula a little bit to calculate any percentile. For example, if you want to calculate 35th percentile, change the formula to L + I *(35% * N-F)/f which is L + I *(0.35 * N-F)/f. Please note that L,I,F & f should reflect that of the group where the percentile falls. You can find this by these steps: 1) Calculate N * 0.35. Lets say N=50 then 50* 0.35 = 17.5. 2) Using cumulative frequency, identify the group where 17.5 falls. 3) Use L,I,F & f for that particular group in the formula L + I *(0.35 * N-F)/f
Mean = sum of all numbers divided by number of numbers you summed. Call numbers a, b, c, d, e, f (a+b+c+d+e+f)/6 = mean
The domain can be any subset of the set of all numbers.
Call the specified numbers a, b, c, d, e, f, and g and the unknown number x. From the problem statement, (a + b + c + d + e + f + g)/7 = 9 [from the definition of "mean" of a set of numbers]. Also (a + b + c + d + e + f + g + x)/(7 + 1) = 9 - 3 = 6, or x/8 = 6 - [(a + b + c + d + e + f + g)/8], or x = 48 - (a + b + c + d + e + f + g).
f(x) = x^{2} is a continuous function on the set R of real numbers, and (-1, 1) is an open set in R, but f(-1, 1) = [0, 1), and [0, 1) is not an open set in R. So, f is not an open function on R.
The GCF of any set of distinct prime numbers is 1.
void main() { float i,j; int a; printf("Enter two numbers : "); scanf("%f%f",&i&j); a=i/j; if(a) printf("%f is greater",i); else printf("%f is greater",j); }