Answer:
On the diagram, construct lines BY, XY and DY.
Now, consider triangles ΔXBY and ΔXDY. We can say that these triangles are congruent by the SAS postulate (they share side XY, XY bisects ∠BYD so ∠BYX=∠DYX, and BY=DY=radius). If they're congruent, their sides are the same so BX=DX.
BX=DX
AB+radius = CD +radius
AB=CD
solution;
a) From the given information,
N = 53,min=42 and max = 129
K = 1+3.322log₁₀53
= 1+3.322(1.724)
= 1+5.728
= 6.728
Number of classes is 7.
b) find the width of the class interval first
width = max – min /classes
= 129-42/7
= 87/7 =12.42
Take the next largest number 13 as class width. Since the minimum value is 42
The lower limit iof the first class could be 40.
So,
The upper limit of the last class is,
40+(13 x 7) = 131
As the upper limit of the last class covers the entire range of values,
It is better to take the lower limit of the first class as 40
The lower limit of the first calss could be 40.
If your in school your teacher taught it: where’s your notes ? haha jk the right answer is B.) bc it’s going over 2 up 3.
Answer:
y = 9/5 x -4
Step-by-step explanation:
first find slope or m
slope is change in y over the change in x
-4-5/0-5 =-9/-5 = 9/5
the (0, -4) is the y intercept or b
Answer:
domt know which one but this may help
Step-by-step explanation:
when you take 38707 into consideration the 3 would be 30,000 the 8 would be 8,000 and the 7 would be 70