Consider the characteristics of the graph. Which statement DOES NOT describe the data set?
2 answers:
Answer:
The correct option is A.
Step-by-step explanation:
In a box plot the left side of box represents Q₁, right side of box represents Q₃ and the line inside the box represents the median.
A line is extended from the left and right side of the box. The left end pint of the line represents the lower limit and right end point of the line represent the upper limit.
From the given it is noticed that the lower limit of the data is 10 and the upper limit of the data is 70. The value of Q₁ is 20 and the value of Q₃ is 40. The median of the data is 30.
The interquartile range (IQR) is the difference between third quartile and first quartile.
Therefore the IQR is 20 and option A does not describes the data set. So, option A is correct.
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Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m