A _____ is a specially designed phone room used to conduct telephone interviewing.

Which approach suggests that a manager believes that employees should be involved in both defining problems and in designing the

morpeh [17]

According to McGregor's Theory Y method, a manager might think that workers ought to be involved in both problem-solving and problem-definition.

One of the theories that has a significant impact on both management and employees is McGregor's theory. Additionally, McGregor's descriptions of two different theories, namely Theory X and Theory Y, are further explained below along with each theory's central tenets.

According to Theory Y, a manager's positive perception of their team problem-solving members will increase employee motivation. Managers erroneously McGregor's Theory Y believe that a decentralized approach that strengthens teamwork, collaboration, and trust can address employee demotivation.

Contrary to Theory X, this theory holds that managers do not believe that problem-solving control motivates workers. The team members must be motivated by McGregor's Theory Y meeting their needs for social interaction, self-actualization, and self-esteem.

Learn more about McGregor's Theory Y here

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6 0

1 year ago

What does it mean for a savings account to have a minimum balance?

Fittoniya [83]

Answer:

D

Explanation:

A savings account is an account owned by an individual at a bank. He keeps his money there, and earn interest at the end of the month for his savings.

A minimum balance is sometimes required by the banks for a savings account, customers are expected to keep such minimum balance always if not, the account will be closed down.

5 0

3 years ago

The Beef-up ranch feeds cattle for midwestern farmers and delivers them to processing plants in Topeka,Kansas and Tulsa, Oklahom

AfilCa [17]

Answer:

a. Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128 .......(1)

3x₁ + x₂ + 3x₃ ≤ 160 ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32 .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256 .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64 ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80 ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16 ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128 ............(8)

x₁ ≤ 15 ...........(9)

x₂≤ 15 ...........(10)

x₃ ≤ 15 ...........(11)

x₁ , x₂, x₃ ≥ 0

b. x₁ = 15 , x₂ = 9.5 , x₃ = 8.5

c. The optimal solution is Z = 79.25

Explanation:

Given - The table is as follows :

Nutrient Feed 1 Feed 2 Feed 3
A 3 2 4
B 3 1 3
C 1 0 2
D 6 8 4

The minimum requirement per cow each month is 4 pounds of nutrient A, 5 pounds of nutrient B, 1 pound of nutrient C, and 8 pounds of nutrient D. However, cows should not be fed more than twice the minimum requirement for any nutrient each month. Additionally, the ranch can only obtain 1,500 pounds of each type of feed each month. Because there are usually 100 cows at the beef-up ranch at any given time, this means that no more than 15 pounds of each type of feed can be used per cow each month.

To find - a. Formulate a linear programming problem to determine how

much of each type of feed a cow should be fed each month.

b. Create a spreadsheet model for this problem, and solve it using

Solver.

c. What is the optimal solution?

Proof -

a.

Let feed 1 per cow per month = x₁

feed 2 per cow per month = x₂

feed 3 per cow per month = x₃

Now,

As given, The cost per pound of feeds 1,2, and 3 are $2.00, $2.50, and $3.00, respectively.

So, we have to minimize the cost , Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 4(32)

3x₁ + x₂ + 3x₃ ≤ 5(32)

x₁ + 0x₂ + 2x₃ ≤ 1(32)

6x₁ + 8x₂ + 4x₃ ≤ 8(32)

∴ we get

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128 .......(1)

3x₁ + x₂ + 3x₃ ≤ 160 ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32 .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256 .........(4)

Now, as given

However, cows should not be fed more than twice the minimum requirement for any nutrient each month.

∴ we have

3x₁ + 2x₂ + 4x₃ ≥ $\frac{128}{2}$

3x₁ + x₂ + 3x₃ ≥ $\frac{160}{2}$

x₁ + 0x₂ + 2x₃ ≥ $\frac{32}{2}$

6x₁ + 8x₂ + 4x₃ ≥ $\frac{256}{2}$

and also

No more than 15 pounds of each type of feed can be used per cow each month.

⇒x₁ , x₂, x₃ ≤ 15

So,

The LPP model becomes

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128 .......(1)

3x₁ + x₂ + 3x₃ ≤ 160 ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32 .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256 .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64 ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80 ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16 ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128 ............(8)

x₁ ≤ 15 ...........(9)

x₂≤ 15 ...........(10)

x₃ ≤ 15 ...........(11)

x₁ , x₂, x₃ ≥ 0

b.)

We use simplex method calculator to solve this LPP Problem

we get

x₁ = 15 , x₂ = 9.5 , x₃ = 8.5

c.)

The optimal solution is Z = 79.25

4 0

2 years ago

On January 1 st 2012, Everhart Corporation, a calendar year company issues $100,000, 5%, 5-year bonds dated January 1, 2012. The

Alborosie

Answer:

Interest expense 2894.7 debit

discount on Bonds Payable 394.7 credit

cash 2500 credit

Interest expense 2906.55 debit

discount on Bonds Payable 406.55 credit

interest payable 2500 credit

Explanation:

We have to solve for the 2013 year which is one year after the issuance ofthe bonds.

We solve for the bond issuance price and then, we construct the bonds schedule and take the numbers from period 3 and 4.

Issuance proceeds: present value fo the coupon payment and maturity at market rate:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 2,500.000

time 10

rate 0.03

$2500 \times \frac{1-(1+0.03)^{-10} }{0.03} = PV\\$

PV $21,325.5071

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity 100,000.00

time 10.00

rate 0.03

$\frac{100000}{(1 + 0.03)^{10} } = PV$

PV 74,409.39

PV c $21,325.5071

PV m $74,409.3915

Total $95,734.8986

Now we will calcautlethe interest expense by multiplying carrying value by the market value and sutract from the cash outlay to determinate the amortization on the bonds.

7 0

2 years ago

Suppose that the economy is characterized by the following behavioral equations:

bezimeni [28]

Answer:

1a. 1400 1b.1230 1c. Equal to

Explanation:

C= 170+0.7(yd)

Y= C+I+G

=170+0.7(Y-100)+170+150

=170+0.7Y-70+170+150

Y =1400

Z=170+0.7(1400-100)+170+150

=170+910+170+150

=1400

It is equal as evident above

4 0

3 years ago