The histograms below show the mass (in kilograms) of the offensive linemen on the University of Mount Union and the University of Alabama football teams. Which pieces of information can be gathered from these histogramsChoose all answers that apply:6) The linemen who play for Mount Union have more mass on average. All of the Alabama linemen have more mass than all of the Mount Union linemen.7 © None of the above

Option C is correct.

None of the options is true.

Step-by-step explanation:

The histograms for the question is missing, so, it was obtained online and is attached to this solution of the question.

To check which statements are true, we first find the average using the midpoint technique.

The midpoint for each of the ranges is presented below

110 to 120 = 115

120 to 130 = 125

130 to 140 = 135

140 to 150 = 145

150 to 160 = 155

Sum of weights for Alabama linemen using the midpoint technique

= (125×4) + (135×8) + (145×4) + (155×2)

= 2470 kg

Average = (2470) ÷ (4+8+4+2)

= 137.22 kg

Sum of weights for Mount Union linemen

= (115×5) + (125×4) + (135×6) + (145×2)

= 2,175 kg

Average = (2,175) ÷ (5+4+6+2)

= 127.94 kg

So, it is evident that the average weight of Mount Union linemen isn't more than that of Alabama linemen.

Statement 2

All of the Alabama linemen have more mass than all of the Mount Union linemen

This is evidently not true because from the histogram, some linemen from Alabama have masses between 120 and 130, 130 and 140 & there are linemen from Mount Union that have masses between 130 and 140 & 140 and 150, so, it isn't right to say that all of the Alabama linemen have more mass than all of the Mount Union linemen.

So, none of the options is correct, option C.

Hope this Helps!!!

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

the answer is 1030 it's very simple math.