Answer:
The answer is the option D
Step-by-step explanation:
we have
Multiply by 
both sides
Rewrite
Answer:
upper right
Step-by-step explanation:
The net for a figure is a representation of all of the figure's faces. The polygons in the net are joined on an edge corresponding to a common edge between the faces they represent.
The given figure is a rectangular prism. Its faces are only rectangles. All nets that have triangles in the net can be eliminated from consideration.
The appropriate choice is the one attached.
The question is incomplete:
1. A cosmetologist must double his/her salary before the employer con realize any profit from his/her work, Miss, Mead paid Miss, Adams $125,00 per week to start.
2. Miss. Mead pays Miss. Brown $125.00 per week. How much money must Miss. Brown take in for services if Miss. Mead is to realize $50.00 profit on her work? (Conditions on salary are the same as in problem 1)
ODS
a. $275.00 b. $325.00 c. $250.00 d. $300.00
Answer:
d. $300.00
Step-by-step explanation:
Given that a cosmetologist must double her salary before the employer can realize any profit from his/her work, for Miss. Mead to realize $50.00 profit on her work, you would have to determine the amount that doubles the salary of the cosmetologist and add the $50 needed as profit:
Salary= $125*2=$250
$250+$50= $300
According to this, the answer is that for Mead to realize $50.00 profit on her work, Miss. Brown must take $300.
NO.2 answer is 5088.052869
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
