Answer:
2
Step-by-step explanation:
Given the data : 29, 2, 28, 30, 26, 31
Outlier ;
Lower :Q1 - (1.5 * IQR)
Upper : Q3 + (1.5 * IQR)
Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range
Using calculator :
Q1 = 26
Q3 = 30
IQR = (Q3 - Q1) = 30 - 26 = 4
Lower : 26 - (1.5 * 4) = 20
Upper : 30 + (1.5 * 4) = 36
Hence, the number in the given data which falls outside the range is 2
2. Definition of perpendicular lines
3. Given
4. SD and DS are congruent
5. HL theorem
Answer:
It is an acute triangle.
Step-by-step explanation:
It would be much easier for you to see where I'm getting at if you draw a diagram.
Assume this triangle is labelled ABC where a = 14, b = 21, and c = 16.
cos A = (16^2 + 21^2 - 14^2)/(2(16)(21))
∠A = cos^-1(0.82291...)
∠A ≅ 34.6°
cos B = 16^2 + 14^2 - 21^2)/(2(16)(14))
∠B = cos^-1(0.024...)
∠B ≅ 56.8°
∠C = 180 - (∠A + ∠B)
∠C ≅ 56.8°
Since there is no right-angle nor is there an obtuse angle,
Therefore the triangle is an acute triangle.
Answer:
5m
Use the Pythagorean theorem: base =sqrt(13^2-12^2)= sqrt(25) =5
sqrt Is Square root btw
