Answer:
B. C. D are true, A. is false
Step-by-step explanation:
<em>The question is attached</em>
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<u>Answer options:</u>
A. Players at school 1 typically spent more time in the weight room than players at school 2
- False, as per plot boxes up to 75% of players from school 1 spend up to 10 hours but up to 75% of players from school 2 spend up to 12 hours in the weight room
B. The middle half of the data for school 1 has more variability than the middle half of the data for school 2
- True. 4 hours variation for school 1 vs 3 hours variation for school 2
C. The median hours spent in the weight room for school 1 is less than the median for school 2 and the interquartile ranges for both schools are equal
- True as it is 8 hours for school 1 and 9 hours for school 2
D. The total number of hours spent in the weight room for players at school 2 is greater than the total number of hours for players at school 1
- True. Taking into account a sum of 5 numbers of the plot box. For School 1 it is: (3+4+8+10+16)= 41 and for school 2 it is: (3+6+9+12+14) = 44. Totals will have similar ratio.
4x^2 -28x + 49=0
- move -49 to the other side
x^2-7x+49/4=0
- take out the 4 because you cannot complete the square with a coefficient in front of x^2
(x^2-7x +(7/2)^2) + 49/4 - (7/2)^2 = 0
- complete the square
4((x-7/2)^2)= 0
- bring the 4 back and you’re done.
hope this helped :)