How does distribution add value to goods and services being sold,
including intellectual property?

Answers

Answer 1

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer 2

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33


Related Questions

A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.

Answers

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views

Answers

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?

a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree

Answers

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?

Answers

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]

plugging values we get

245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter

Answers

Answer:

shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = [tex]\frac{2\pi NT}{60}[/tex]  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 )  = 9.554 KN-M

next

[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]

hence ; T = Z[tex]_{p} *Z[/tex]

attached below is the remaining part of the solution

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