Given Set \(M:\ 0,\ 1,\ 3,\ 4,\ 7\) and Set \(N:\ 1,\ 2,\ 3,\ 4,\ 5,\ 9\). Calculate the mean and mean absolute deviation for both sets. Then, answer the following questions.

1. How many data points in Set M are closer than one mean absolute deviation from the mean of Set M?
2. How many data points in Set N are further than two mean absolute deviations from the mean of Set N?
3. Find the difference between the two means. How does this difference relate to the mean absolute deviation of each set? For example, is it the same as the mean absolute deviation, twice as much, half as much, etc?

1. 5

2. 6

3 . difference between the two means is 2,5,9

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)