Answer: A. Paul’s data has a bigger overall spread than Sally’s data
The range of Paul’s Data: 28-12=16
The range of Sally’s Data: 30-15=15
16>15. Therefore Paul’s data has a bigger overall spread than Sally’s
Why B is wrong: Remember that the interquartile range is the “box” range.
Paul’s IQR: 26-15=11
Sally’s IQR: 28-18=10
11<10 is false, so B is false.
Why C and D is wrong: The middle line in the IQR represented in each box and whisker plot is the median. Paul’s median is not equal nor greater than Sally’s median. 18=24 18<24 is false. Therefore, C and D are false.
Answer:
the answer is 848500 centimetres
Step-by-step explanation:
we have to multiply the value by 100.
hope it helps.
Answer:
Square root of 16 is not 16
Answer:
Q : 3
10x - 11 = 120 - 11 = 109°
3x - 2 = 36 - 2 = 34°
3x + 1 = 36 + 1 = 37°
Q ; 2
3x - 5 = 27 - 5= 22°
7x + 5 = 63 + 5 = 68°
And 90°
Q1 :
∠1 = 92°
∠2 = 42°
∠3 = 113°
Step-by-step explanation:
Solution for Q : 3
As the angle of all three is given as ,
10x - 11
3x - 2
3x + 1
We know sum of all the three angles of triangle = 180 °
So, (10x - 11) + (3x - 2) + (3x + 1) = 180°
Or, 16x - 12 = 180°
Or 16x = 192°, So , x = 12
So, all three angles are 10x - 11 = 120 - 11 = 109°
3x - 2 = 36 - 2 = 34°
3x + 1 = 36 + 1 = 37°
Solution for Q - 2
Given angles are
3x - 5
7x + 5
90°
We know sum of all the three angles of triangle = 180 °
so ,(3x - 5) + (7x + 5) + 90 = 180°
or 10x = 180 - 90 = 90°
SO, x = 9°
SO, all the three angles are 3x - 5 = 27 - 5= 22°
7x + 5 = 63 + 5 = 68°
And 90°
Solution for Q : 1
From,
the shown fig it is clear that
The ∠2 = 42° (<u> opposite vertical angles</u> )
so, in left triangle
50° + ∠2 + ∠1 = 180°
Or, 50° + 42° + ∠1 = 180° ( sum of all angles of triangles = 180°)
Or, ∠1 = 92°
Again
From right figure triangle
∠2 + 25° + ∠3 = 180°
Or, 42° + 25° + ∠3 = 180
Or, ∠3 = 113°