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Scorpion4ik [409]
2 years ago
14

Find the mistake and correct it

Mathematics
1 answer:
Murljashka [212]2 years ago
6 0

Answer:

i believe its the second part thats wrong  

Step-by-step explanation:

1 Distribute

−5(−2+10)=20

10−50=20

2 Add 50 to both sides of the equation

10−50=20

10−50+50=20+50

3 Simplify

Add the numbers

10x−50+50=20+50

10=20+50

Add the numbers

10x =20 + 50

10x=10

4 Divide both sides of the equation by the same term

10x = 70

10x   70

---- = ----

10      10

5 Simplify

Cancel terms that are in both the numerator and denominator

10x   70

---- = ----

10      10

      70

x  = ----

       10

   

        70

x  = ----

       10  

x=7

Solution

= 7

hopes it helps and list brain list plz

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Answer:

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b. $0.93

c. 0.98

d. 2 workers

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a. Given that:

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  • 1 computer : 2 workers : inventory 200 items per hour
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The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

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Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98

Hence, the cost of inventorying a single item is $0.98

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Take any number less than 235(say 234) as the inventory units:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

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