Question

the median and the mode of the data : x,y,z,3,19,5,11,3,19,5,11 are y,x where 3<x<y if the range of the data is 17 and x,y are prime no. then (x+y+z) equals to ●30●32●31●34​

Answer 1

Answer:

32

Step-by-step explanation:

x,y,z,3,19,5,11,3,19,5,11

here,

median is = y & mode = x

and the range of the data is 17

first ,

range of the data = highest value in data -  lowest value in data

but in this data  highest value in data -  lowest value in data  

= 19 - 3

16

but range is 17

and y and x are median and mode so z must be the highest value in data

range = z -3

17 = z-3

z = 20

now for mode and median arrange the data in ascending

3 , 3 ,5 ,5 , x , y , 11 , 11 , 19 , 19 , 20

y is median so its is in the middle of it

now x and y are prime

and 3<x<y

so the value of the y is between 5 and 11

so y = 7  because 7 is the prime between 5 and 11

we can not take the value of y is 5 or 11 because all the numbers given in data are twice if we take the value of y 11 or 5 the y will be the moe

so,

y=7

now for value of x which is mode

3<x<y

3<x<7

so x= 5 because only 5 is the value between 3 and 7 which can be a mode

so (x+y+z)

=5 + 7 +20

=32

Answer 2

Answer:

32 is the correct answer