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zzz [600]
3 years ago
6

Which of the following is NOT a recommended characteristic for incident objectives?A. Stated in broad terms to allow for flexibi

lityB. Measurable and attainableC. In accordance with the Incident Commander's authoritiesD. Includes a standard and timeframe
Computers and Technology
1 answer:
docker41 [41]3 years ago
6 0

Answer:

A. Stated in broad terms to allow for flexibility.

Explanation:

Incident command are controls or directives needed in production or in an organization. A few characteristics not incident command objectives are, all projects and targets must be measurable and attainable, aii hand on board must function in accordance with the incident commander's authorities, and every finished project must meet company quality standards and deadlines. The incident commander's authorities are fixed for a particular sector and is rigid.

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List two panels that allow you to adjust the properties of your titles.
antiseptic1488 [7]

Answer:

1. The tools panel

2. The actions panel

Explanation:

The tools panel contains properties that allow for text and object creation. The actions panel contains properties that allow for the alignment and distribution of titles.

Under the tools panel, we can find properties like; line, arc, an arrow for selection, different shapes like rectangle and the clipped corner, vertical type, vertical area type, vertical path type, etc. Under the actions panel, properties like; align, center, and distribute can be found.

8 0
3 years ago
What is an indication that malicious code is running?
tatyana61 [14]

Answer:

If you are running Windows check task manager and see if any background process if running if there is a background process of command prompt or powershell, there is a chance that there is a backdoor running in your system, you can just right click and end the task it should just kick the hacker off the server.

please give brainliest

7 0
3 years ago
Find the double word-length 2's complement representation of each of the following decimal numbers:a. 3874
miss Akunina [59]

Answer:

-3874₁₀ = 1111 1111 1111 1111 1111 1111 1101 1110₂

Explanation:

2's complement is a way for us to represent negative numbers in binary.

To get 2's complement:

1. Invert all the bits

2. Add 1 to the inverted bits

Summary: 2's complement = -N = ~N + 1

1. Inverting the number

3874₁₀ = 1111 0010 0010₂

~3874₁₀ = 0000 1101 1101₂

2. Add 1 to your inverted bits

~3874₁₀ + 1 = 0000 1101 1101₂ + 1

= 0000 1101 1110₂

You can pad the most signigicant bits with 1's if you're planning on using more bits.

so,

12 bits                          16 bits

0000 1101 1110₂  = 1111 0000 1101 1110₂

They asked for double word-length (a fancy term for 32-bits), so pad the left-most side with 1s' until you get a total of 32 bits.

           32 bits

= 1111 1111 1111 1111 1111 1111 1101 1110

7 0
3 years ago
How many microprocessors will a small computer have
Arte-miy333 [17]
Well, most of the new Pc's today have dual-core central processors (CPU). They only have one chip with two complete microprocessors on it, both of them share one path to memory and peripherals. But if it's a high-end hardcore gaming Pc, you could have two processor chips with 4 microprocessors on each.
4 0
4 years ago
Read 2 more answers
An upper-layer packet is split into 10 frames, each of which has an 80% chance of arriving undamaged. If no error control is don
Arisa [49]

Answer: The message must be sent 9.313 approximately 9 times to get the entire data through.

Explanations: To find how many times the message must be sent on average to avoid error control in data link later.

E = 1/P

E = The average number of attempt before successful transmission.

P= Total probability of transmission without error.

STEP1 : FIND TOTAL PROBABILITY;

Since it each frame has a probability of 80% to be successful.

For each frame p = 80/100 = 0.8

For the 10 frame; total probability

P= (0.8)^10 = 0.1074

STEP2: FIND THE AVERAGE NUMBER OF OF TRIAL BEFORE A SUCCESSFUL TRANSMISSION WITHOUT ERROR;

Using equation above

E = 1/P

E= 1 ÷ 0.1074 = 9.313

Therefore they must be an average of 9.313 approximately 9 trials before a successful transmission without error.

5 0
3 years ago
Read 2 more answers
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