The average of x, 1, 2, 3 is 4, find the average of x, 1 and 2.
my answer was 3 is it correct ?
Not quite. Can you show your work so that we can pinpoint any errors you have made? To start, you know the average is 4. So, the sum of all the numbers divided by the number of numbers is equal to 4:
$\displaystyle 4 = \frac{x + 1 + 2 + 3}{4}$
You want to find the average of x, 1, and 2 (the sum of these numbers divided by 3 since there are 3 numbers now):
$\displaystyle \mbox{Average} = \frac{x + 1 + 2}{3}$
Can you think of a way to use the first equation to get the average in the second equation?
Hello,
If the average of x, 1, 2 and 3 is 4, you can write :
$\displaystyle \frac{x+1+2+3}{4}=4$ (1)
And you're looking for $\displaystyle \frac{x+1+2}{3}$
From (1), we have : $\displaystyle x+1+2+3=16$
So $\displaystyle x+1+2=13$
Then $\displaystyle \underbrace{\frac{x+1+2}{3}}_{\text{average \ of \ x, \ 1, \ 2}}=\frac{13}{3}$
Hello, sri340
Are you sure you know what an average is?The average of x, 1, 2, 3 is 4.
Find the average of x, 1 and 2.
my answer was 3 is it correct ? . . . . no
"The average of x, 1, 2, 3 is 4."
This means: .$\displaystyle \frac{x + 1 + 2 + 3}{4} \:=\:4$
. . Hence: .$\displaystyle x + 1 + 2 + 3 \:=\:16\quad\Rightarrow\quad x + 1 + 2 \:=\:13$
Divide by 3: . $\displaystyle \frac{x + 1 + 2}{3} \:=\:\frac{13}{3}$
. . Therefore, the average of $\displaystyle x,1,2\text{ is: }\:\frac{13}{3}$