Pls help me! (question is in the picture)
2 answers:
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Answer:
Step-by-step explanation:
<em>Given:</em>
Mn is diameter of circle having centre O
and BD = OD,
<em><u>To prove that:</u></em>
<u></u>
<em>Solution:</em>
Join the points O and B and draw OB,
On joining the line,
in ∆OCD and ∆OBD,
OC =OB → (Radius of same circle)
BD =CD → (from given)
OD =OD → (Common side in both the triangles)
Hence ∆OCD and ∆OBD are congruent from SSS property.
so we can say that,
Consider above prove as statement A
Corresponding angles of congruent traingle.
in ∆ OAB,
OA = OB (radius of same circle)
hence ∆OAB is an isosceles traingle.
We know that opposite angle of isosceles traingle are always equal. hence,
Consider above prove as statement B
From Statement A & B we can say that
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Answer:
The interquartile range of the data is 12
Step-by-step explanation:
Data : 14, 22, 19, 21, 30, 32, 25, 15, 16, 27, 28 12 16 22 28
Data in ascending order : 12 ,14 ,15 ,16 ,16 ,19 ,21 ,22 ,22 ,25 ,27 ,28 ,28 ,30 ,32
Lower quartile :12 ,14 ,15 ,16 ,16 ,19 ,21
Q1 is the median of lower quartile
n = 7
Median ==16
Upper quartile :22 ,25 ,27 ,28 ,28 ,30 ,32
Q3 is the median of upper quartile
n = 7
Median ==28
IQR = Q3-Q1=28-16= 12
Hence the interquartile range of the data is 12