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Mamont248 [21]
3 years ago
9

E-mail messages, instant messages (IMs), or text messages sent and/or received within an organization a. are not included on a r

ecords retention schedule. b. may be considered as records or nonrecords. c. original messages are retained in the e-mail system indefinitely. d. do not reduce mailing costs.
Computers and Technology
1 answer:
Elanso [62]3 years ago
6 0

Answer:

b. may be considered as records or non records

Explanation:

Message in an organization and even instant messages (IMs) can be considered as records or nonrecords, this depends on the content, nowadays in a company can be used apps like social media, important information can be stored between customers or partners, this no reduces any cost and the message can be deleted depends on the server settings.

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The name for the instructions you write to a computer in a program
aliina [53]

Answer:

Code

Explanation:

The code is instructions that you can write yourself or download from online

4 0
2 years ago
Read 2 more answers
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
You are given a class named Clock that has one int instance variable called hours.
Vlad [161]

Answer:

public Clock(int hours) {

       this.hours = hours;

   }

Explanation:

In Java programming language, Constructors are special methods that are called to initialize the variables of a class when a new object of the class is created with the new keyword. Consider the complete code for the class below;

<em>public class Clock {</em>

<em>    private int hours;</em>

<em>    public Clock(int hours) {</em>

<em>        this.hours = hours;</em>

<em>    }</em>

<em>}</em>

In this example above,  an object of this class can created with this statement Clock myclock = new Clock(6); This is a call to the constructor and passes a parameter (6) for hours

7 0
3 years ago
1. Assume that word is a variable of type String that has been assigned a value . Assume furthermore that this value always cont
s2008m [1.1K]

Answer:

1.word = "George slew the dragon"

startIndex = word.find('dr')

endIndex = startIndex + 4

drWord = word[startIndex:endIndex]

2. sentence = "Broccoli is delicious."

sentence_list = sentence.split(" ")

firstWord = sentence_list[0]

Explanation:

The above snippet is written in Python 3.

1. word is initialized to a sentence.

Then we find the the occurence of 'dr' in the sentence which is assign to startIndex

We then add 4 to the startIndex and assign it to endIndex. 4 is added because we need a length of 4

We then use string slicing method to create a substring from the startIndex to endIndex which is assigned to drWord.

2. A string is assigned to sentence. Then we split the sentence using sentence.split(" "). We split based on the spacing. The inbuilt function of split returns a list. The first element in the list is assigned to firstWord. List uses zero based index counting. So. firstWord = sentence_list[0] is use to get first element.

4 0
3 years ago
FREE POINT IF YOU ADD ME AS A FRIEND <br> XOXOANAI04
scoray [572]

Answer:

hfufbrdubedyg gxnsbetxyfbrdybeuxuabrbxybsbcyebdydbehdhdhdhcubdenudnejwjw7xtrwhajaobcbdggevejaveyxbwbgzfybaegnesujwbghene8hav ejuzvwbfhuh bsndndbchchychdbruxnejxne7hdrhxhehnxurnxhhbehdndjndhd

8 0
3 years ago
Read 2 more answers
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