" 7. Mr Gerry’s class has 19 students Ms Vernon’s class has 21 students and Mr Singh’s class has 23 students Whose class has a c

To calculate the mean absolute deviation of
1,2,3,4,5,6,7
, we start by finding the mean;
 (1+2+3+4+5+6+7) =28/7
                               = 4
. Next, we determine the absolute differences of each data point from the mean (x-μ)
 = -3,-2,-1,0,1,2,3
. The absolute values are 3,2,1,0,1,2,3
. Now we compute the mean of these absolute differences,
3+2+1+0+1+2+3 = 12
                           = 12/7
                           = 1.7143
. Thus, the mean is 4, and the Mean absolute deviation comes out to be 1.7143

therefore



and you are probably aware of the amount.

Answer:

B. y - 2 = 1/3(x - 3)

<span>9 ten thousandths equals 0.0009, which can be expressed as 9 × 10 to the power of -4</span>

A.
I hope this information is useful to you.