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katrin [286]
3 years ago
15

How to make multiple invitations on microsoft word?

Computers and Technology
1 answer:
Anton [14]3 years ago
8 0
Go to the share with bar and type in whoever's email address that you want to share the document with
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There are no breakpoints in the "Access Customer Account" subpage however there is an error. What will happen if you choose to s
blsea [12.9K]

Answer:

Move to the breakpoint at "Get Customer Details" stage.

Explanation:

This question is incomplete, we need a diagram and four options to resolve this questions, I attached the diagram and the options.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the breakpoint at "Get Customer Details" stage.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the "Recover1" stage.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the stage containing the error on the "Access Customer Account" page.

- The process will work all stages in the "Access Customer Account" page until the error is thrown and then focus would move to the "Exception1" stage.

In this case, we are going to run an extra from subpage testing in <u>Process studio</u>. Process studio is a tool where we can test these processes.

The process always it's going to work all stages in the "Access Customer Account" we're going to receive an interruption when the error is activated, moving the focus to the Get Customer Details stage.

There is an error in the stage Access Customer Account, if we run the processes with the Go button, the focus should move into the Recover1 stage, and should show the error.

But in this particular example, we must use the STEP OUT BUTTON, What means this?

The process it's going be executed, but won't show any error or message, because we have used the STEP OUT BUTTON.

If we use the STEP OUT BUTTON, the process should end but in this case, we have a breakpoint in Get Customer Details stage, for that the focus and the process will end in Get Customer Details stage and not at the end or at the Recover1 stage.

8 0
3 years ago
Anyone knows the answer for 6.1.4 Happy Birthday! codehs
Neko [114]

cvm is good cvm is great

7 0
2 years ago
What is the advantage of using CSS?
zhannawk [14.2K]
Your answer would be B.) "It streamlines the HTML document."
8 0
3 years ago
Read 2 more answers
We want to implement a data link control protocol on a channel that has limited bandwidth and high error rate. On the other hand
Korolek [52]

Answer:

Selective Repeat protocols

Explanation:

It is better to make use of the selective repeat protocol here. From what we have here, there is a high error rate on this channel.

If we had implemented Go back N protocol, the whole N packets would be retransmitted. Much bandwidth would be needed here.

But we are told that bandwidth is limited. So if packet get lost when we implement selective protocol, we would only need less bandwidth since we would retransmit only this packet.

6 0
3 years ago
5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0
3 years ago
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