Refer to the figure below to complete the following problem.

Normal text

Gnoma [55]

Answer:

your mom

Step-by-step explanation:

3 0

3 years ago

Imagine you have some workers and some handheld computers that you can use to take inventory at a warehouse. There are diminishi

Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

1 computer : 1 worker : inventory 150 items per hour
1 computer : 2 workers : inventory 200 items per hour
1 computer : 3 workers : inventory 220 items per hour
1 computer : 4+ workers : fewer than 235 items per hour
Cost: $100 per computer ; $25 per worker

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.

#The cost of inventorying a single item by one worker is:

$Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03$

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

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

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.

4 0

3 years ago

2. Solve the system of equations. Write your answer in parametric vector notation. X, +5x2 + 4x3 + 3x4 +9xs = 18 2x, +6x2 + 4x3

ratelena [41]

Answer:

The system is inconsistent.

Step-by-step explanation:

The system is

$x_1+5x_2+4x_3+3x_4+9x_5=18\\2x_1+6x_2+4x_3+6x_4+6x_5=28\\3x_1+7x_2+4x_3+10x_4+x_5=41\\4x_1+6x_2+2x_3+7x_4+4x_5=29$

The associated matrix to the system is

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\2&6&4&6&6&28\\3&7&4&10&1&41\\4&6&2&7&4&29\end{array}\right]$

Now we use row operations to find the echelon form of the matrix:

1. We substract to row 2, two times the row 1.

We substract to row 3, three times the row 1.

We substract to row 4, four times the row 1 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&-4&-4&0&-12&-8\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

2. We multiply the second row of the preview step by -1/4. We obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&-8&-8&1&-26&-13\\0&-14&-14&-5&-32&-43\end{array}\right]$

3.

We add to row 3, eight times the row 2.

We add to row 4, fourtheen times the row 2 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&-5&10&-155\end{array}\right]$

4. We add to row 4 of the preview step, five times the row 3 and obtain the matrix

$\left[\begin{array}{cccccc}1&5&4&3&9&18\\0&1&1&0&3&2\\0&0&0&1&-2&3\\0&0&0&0&0&-140\end{array}\right]$

Using backward substitution we have that

$0x_5=-140$ , then $0=-140$ and this is absurd. Then The system is inconsistent.

4 0

4 years ago

for a special concert by a popular band, the concert hall owner decided to charge $20 per ticket for the first "r" rows and $15

Alecsey [184]

R = 16 according to the problem, so the total amount of income will be:
$20*16 + $15*(40 - 16)
= $320 + <span>$15*(24)
= </span>$320 + <span>$360
= $680
that is the income, but the expenses are $275 so the total revenue is the subtraction of those two
total revenue = $680 - $275
= $405</span>

7 0

3 years ago

A circular swimming pool has a radius of 28ft. There is a path all the way around the pool that's 4ft wide. A fence is going to

Natalka [10]

Answer:

Therefore 200.96 ft.of fencing are needed to go around the pool path.

Step-by-step explanation:

Given, a circular swimming pool has a radius of 28ft. There is a path all the way around the pool. The width of the path 4 ft.

The radius of the outside edge the pool path is

= Radius of the pool + The width of the path

= (28+4) ft

= 32 ft.

To find the length of fencing, we need to find the circumference of outside the pool path.

Here r= 32 ft

The circumference of outside edge of the pool path

= $2\pi r$

$=(2\times 3.14 \times 32) ft$

=200.96 ft.

Therefore 200.96 ft.of fencing are needed to go around the pool path.

3 0

3 years ago