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Sagot :
If q is an odd number and the median of q consecutive integers is 120, then the largest of these integers is option (A) (q-1) / 2 + 120
The number q is an odd number
The median of q consecutive integers = 120
Consider the q = 3
Then three consecutive integers will be 119, 120, 121
The largest number = 121
Substitute the value of q in each options
Option A
(q-1) / 2 + 120
Substitute the value of q
(3-1)/2 + 120
Subtract the terms
=2/2 + 120
Divide the terms
= 1 + 120
= 121
Therefore, largest of these integers is (q-1) / 2 + 120
I have answered the question in general, as the given question is incomplete
The complete question is
if q is an odd number and the median of q consecutive integers is 120, what is the largest of these integers?
a) (q-1) / 2 + 120
b) q/2 + 119
c) q/2 + 120
d) (q+119)/2
Learn more about median here
brainly.com/question/28060453
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