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REY [17]
3 years ago
10

Help me find the right answer please.

Computers and Technology
1 answer:
timama [110]3 years ago
5 0

Answer:

I'm not sure

Explanation:

I'm good at math but I do not see any numbers to work with. SORRY

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g You and your friend both send documents to the printer, but the printer can only print one document at a time. What does it do
Sedbober [7]

Answer:

It places the document in a buffer

Explanation:

Since the printer can only print a document at a time, the other document is going to be placed in a buffer.

The print buffer can be described as a location in memory that has the function of holding data that is going to be sent to a computers printer. A print job such as in this scenario may remain in the buffer because another document is to be printed first.

Sometimes a document could be put in the buffer because the computer is waiting for the printer to respond. This is just another reason why documents are placed in a buffer.

5 0
3 years ago
Jacob is preparing a presentation on the health and social advantages of taking up a sport at an early age. Most of the slides i
NARA [144]

Answer:

It's C

Explanation:

On edgenuity

4 0
3 years ago
Given that a computer required​ service, what is the probability that it was a​ laptop? Given that the computer required​ servic
Greeley [361]

Complete Question:

A campus bookstore sells both types and in the last semester sold 56% laptops and 44% desktops. Reliability rates for the two types of machines are quite different, however. In the first year, 5% of desktops require service, while 15% of laptops have problems requiring service.

Given that a computer required service, what is the probability that it was a laptop?

Answer:

Probability = 0.084

Explanation:

Given

Laptops = 56%

Desktop = 44%

Service Required (Laptop) = 15%

Service Required (Desktop) = 5%

Required

Determine the probability that a selected computer is a laptop and it requires service.

The question tests our knowledge of probabilities using "and" condition.

What the question requires is that, we calculate the probability of selecting a LAPTOP that REQUIRES SERVICE

Note the capitalised words.

This will be calculated as follows:

Probability = P(Laptop) and P(Service Required (Laptop))

[Substitute values for P(Laptop) and P(Service Required (Laptop))]

Probability = 56% * 15%

[Convert to decimal]

Probability = 0.56 * 0.15

Probability = 0.084

8 0
3 years ago
Which method(s) must a serializable class, implement?
Cloud [144]

Answer: None of the given option is correct

Explanation:

A serializable class is implemented, when an object is serializable interface. If you want to serialize one of your classes, then the class must implemented in the Serializable interface. To implement the serializable interface, convert a class into the series of bytes and when serializable object might reference your class. Serializable classes are useful when you wanted to persist cases of the class or send them over wire.

5 0
3 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
2 years ago
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