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Which is correct regarding the statement: “If x is an odd integer, then the median of x, x + 2, x + 6, and x + 10 is an odd number”

Respuesta :

apologiabiology
any odd number can be represened as 2n+1 where n is a whole number

the median is the  number in the middle when the numbers are arranged in increasing order
if there are an even number of numbers in the set, take the average of the middle 2


we have x,x+2,x+6,x+10
there are 4 numbers, so the median will be the avergage between x+2 and x+6 which is [tex]\frac{x+2+x+6}{2}=\frac{2x+8}{2}=x+4[/tex]
now, if x is odd, then replace x with 2n+1

(2n+1)+4=
2n+1+4=
2n+5=
2n+2(1)+1
even+even+odd=odd
the statement is true
x is odd
x+2 is also odd (eg: x = 3, so x+2 = 3+2 = 5)
x+6 is also odd for similar reasons above
x+10 is also odd

We have four values. The median is found in the middle, so exactly halfway between x+2 and x+6

Add up x+2 and x+6 to get
(x+2)+(x+6) = 2x+8

Then divide that result by 2. In other words, divide everything by 2
2x/2+8/2 = x+4

The result x+4 is odd since adding any even number to an odd number is going to result in an odd value

So the claim is true

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Alternative method:
x is odd so x = 2k+1 where k is any integer
x+2 = 2k+1+2 = 2k+3
x+6 = 2k+1+6 = 2k+7
x+10 = 2k+1+10 = 2k+11

Add up the two middle terms
(x+2)+(x+6) = (2k+3)+(2k+7) = 4k+10

Divide that result in half
4k+10 ---> 4k/2 + 10/2 = 2k+5 = 2k+4+1 = 2(k+2)+1

The expression 2(k+2)+1 fits the form 2m+1 where m = k+2 is an integer

So 2(k+2)+1 is odd making the median to be odd. 

This further helps support why the claim is true

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