Answer:
try letter B
Step-by-step explanation:
and can you be my friend
Answer:
a) 28
b)
Class Limits Class Boundaries Midpoint f Rf Cf
10 - 37 9.5-37.5 23.5 7 0.096 7
38 - 65 37.5-65.5 51.5 25 0.342 32
66 - 93 65.5-93.5 79.5 26 0.356 58
94 - 121 93.5-121.5 107.5 9 0.123 67
122 - 149 121.5-149.5 135.5 5 0.068 72
150 - 177 149.5-177.5 163.5 0 0.0 72
178 - 205 177.5-20.5 191.5 1 0.014 73
Total 73 1.0
c)
The histogram is in attached file.
Step-by-step explanation:
a)
The width of class interval=range/desired number of classes
The width of class interval=(maximum-minimum)/7
The width of class interval=(200-10)/7
The width of class interval=190/7
The width of class interval=27.14=28 (rounded to next whole number)
Part(b) and part (c) is explained in attached word document.
Answer:
0
1
Step-by-step explanation:
First question:
You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.
The sine function can never equal 2, so there is no triangle in this case.
Answer: no triangle
Second question:
You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.
One triangle exists for sure. Now we see if there is a second one.
Now we look at the supplement of angle C.
m<C = 52.5°
supplement of angle C: m<C' = 180° - 52.5° = 127.5°
We add the measures of angles B and the supplement of angle C:
m<B + m<C' = 63° + 127.5° = 190.5°
Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.
Answer: one triangle
Answer:
C
Step-by-step explanation:
If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18,
x=(m+9)/2
y=(2m+15)/2
z=(3m+18)/2
the average of x, y, and z
=(x+y+z)/3
=(m+9+2m+15+3m+18)/3
=(6m+42)/3
=2m+14
ans is C