How does the mean absolute deviation (MAD) of the data in set 1 compare to the mean absolute deviation of the data in set 2?
Set 1: 5, 10.7, 2
Set 2: 5, 10, 7, 42
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A)The MAD of set 1 is 10.5 more than the MAD of set 2.
B)The MAD of set 1 is 2.5 more than the MAD of set 2.
C)The MAD of set 1 is 10.5 less than the MAD of set 2.
D)The MAD of set 1 is 2.5 less than the MAD of set 2.

(Please sombody awnser this quickly

the answer is C

Step-by-step explanation:

if you do 5+10+7+2 you get 24 divide by 4 you get 6 mean is 6 for set 1

Set 2 if you do 5+10+7+42 you get 64 divide by 4 and you get 16 16-10.5 you get so 5.5 set is less than set 2 there ya go dude

14 ft 7 inches

step-by-step explanation:

since each model inch corresponds to 25 real inches, 7 model inches will correspond to

7×25" = 175"

at 12" per foot, that is 14 feet, 7 inches.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.