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attashe74 [19]
3 years ago
5

Which communications management practice includes specifying all of the communications systems and platforms that parties will u

se to share information?
A. Equipment Standards
B. Agreements
C. Policy and Planning
D. Standardized Communication Types
Business
1 answer:
artcher [175]3 years ago
4 0

Answer:

The correct answer is B. Agreements.

Explanation:

Agreement is, in Law, the decision taken jointly by two or more persons, or by a board, assembly or court. It is also called a pact, treaty, agreement, convention or resolution taken within an institution (any type of organization or company, public or private, national and international).

It is, therefore, the manifestation of a convergence of wills (decision by consensus) in order to produce legal effects. The main legal effect of the agreement is its obligation for the parties that grant it (Pacta sunt servanda) being born for the same obligations and rights (bilateral or synalagmatic contract), all to the extent provided by applicable law.

The legal validity of an agreement requires that the consent of the grantors is valid and its purpose is true and determined, not out of trade or impossible. Regarding the form of its celebration, oral or written, laws usually require certain formalities that depend on the nature of the obligations agreed upon.

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MC Qu. 93 Schrank Company is trying to decide how... Schrank Company is trying to decide how many units of merchandise to order
ss7ja [257]

Answer:

43,000 units

Explanation:

Given that,

Projected sales in August = 48,000

Projected sales in September = 38,000

Projected sales in October = 58,000

Ending inventory is the 25% of the next month's sales.

The units purchased in the month of September includes the amount of sales in this month and the amount of ending inventory for this month but the amount of beginning inventory excludes from this calculation of purchases.

Ending inventory for September:

= 25% of the next month's sales

= 0.25 × Sales in October

= 0.25 × 58,000

= 14,500

Here, the opening inventory for September is the amount of ending inventory for August.

Opening inventory for September:

= 0.25 × Sales in September

= 0.25 × 38,000

= 9,500

Units to be purchased:

= Sales in September + Ending inventory for September - Opening inventory for September

= 38,000 units + 14,500 - 9,500

= 43,000 units

8 0
3 years ago
Read 2 more answers
In 20X5, Elm Corp. bought 10,000 shares of Oil Corp. at a cost of $20,000. On January 15, 20X6, Elm declared a property dividend
spin [16.1K]

Answer:

b) $24,000

Explanation:

The property dividends are an alternative to cash and stock dividends. Usually because, the firm doesn't have enought cash to give a wealthy dividend so it gives shares of a subsidiary, marketable securities or real state.

They can recognize a gain or sale on the asset, because it will be valued at market value at the time of the distribution. At the time of the distribution, the Oil Corp shares are valued at 24,000 The accounting should represent the reality. This is, dividends were given for 24,000

Adjustment will be made to show the property dividends on 24,000

recognize a gain on oil Corp investmest for 4,000

and the decrease on RE for 24,000

3 0
4 years ago
A buyer purchased a home under an agreement that made the buyer personally obligated to continue making payments under the selle
sammy [17]
<span>A buyer purchased a home under an agreement that made the buyer personally obligated to continue making payments under the seller's existing mortgage. If the buyer defaults and the court sale of the property does not satisfy the debt, the buyer will be liable for making up the difference. The buyer has taken over/assume the sellers mortgage. 

Often, then this happens nothing about the loan changes besides who is paying for it. The payments, interest rates, and loan term agreements stay the same. Someone else just becomes financially responsible for the payments. </span>
8 0
3 years ago
"a negative supply shock causes output to _________ and the price level to _______."
Volgvan
The answers are decrease and increase in that order.
6 0
4 years ago
A point charge q1 = -9.6 μC is located at the center of a thick conducting spherical shell of inner radius a = 2.4 cm and outer
inessss [21]

Answer:

Part 1: The value of x component of electric field  at point P is -1.03 \times 10^7 \frac{ N}{ C}

Part 2: The value of y component of electric field  at point P is 0.

Part 3: The value of x component of electric field  at point R is 0.

Part 4: The value of y component of electric field  at point R is -5.06 \times 10^8 \frac{ N}{ C}.

Part 5: The value of surface density at the outer edge of the shell  is -3.83 \times 10^{-4} C/m^2.

Part 6: None ,The field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of r^2.

Part 7:The fields are equal as the charge on the outer shell does not affect the field on within the shell. (E_2=E_o)

Explanation:

Part 1

As

E = k \frac{ Q}{ r^2} is the Electric field of a point charge

From given data

q_1 = -9.6 \mu C is point charge at the center

q_2 = 1.5\mu C is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

x= 8.4 cm = 0.084 m is the location of P on x

So the value is given as

E_x( P) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.084^2}\right) \\= -1.03 \times 10^7 \frac{ N}{ C}

The value of x component of electric field  at point P is -1.03 \times 10^7 \frac{ N}{ C}

Part 2

As

E = k \frac{ Q}{ r^2} is the Electric field of a point charge

From given data

q_1 = -9.6 \mu C is point charge at the center

q_2 = 1.5\mu C is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

y = 0 cm = 0.0 m is the location of P on y

So the value is given as

E_y( P) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0

The value of y component of electric field  at point P is 0.

Part 3

As

E = k \frac{ Q}{ r^2} is the Electric field of a point charge

From given data

q_1 = -9.6 \mu C is point charge at the center

q_2 = 1.5\mu C is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

x = 0 cm = 0.0 m is the location of R on x

So the value is given as

E_x( R) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0

The value of x component of electric field  at point R is 0.

Part 4

As

E = k \frac{ Q}{ r^2} is the Electric field of a point charge

From given data

q_1 = -9.6 \mu C is point charge at the center

q_2 = 1.5\mu C is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

y =1.2 cm = 0.012  m is the location of R on y

So the value is given as

E_y( R) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.012}\right) \\=-5.06 \times 10^{8} N/C

The value of y component of electric field  at point R is -5.06 \times 10^8 \frac{ N}{ C}.

Part 5

As

\sigma = \frac{ q_{enclosed}}{ 4 \pi r^2} is the Surface charge density

From given data

q_1 = -9.6 \mu C is point charge at the center

q_2 = 1.5\mu C is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

So the value is given as

\sigma_b = \frac{ q_{enclosed}}{ 4 \pi b^2}\\\sigma_b = \frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{ 4 \pi 0.041^2}\\\sigma_b = -3.83 \times 10^{-4} C/m^2

The value of surface density at the outer edge of the shell  is -3.83 \times 10^{-4} C/m^2.

Part 6

<em>None because the field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of </em>r^2.

Part 7

<em>The fields are equal as the charge on the outer shell does not affect the field on within the shell. (</em>E_2=E_o)

5 0
4 years ago
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