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ioda
3 years ago
14

Assume you previously entered these lines of code.

Computers and Technology
2 answers:
antiseptic1488 [7]3 years ago
8 0

Answer:

result = ab + bc and result = a * b - b

Explanation:

That is what the test told me

Stella [2.4K]3 years ago
6 0

Answer:

The correct answer is result = a * b - b / c <em>and</em> result = ab + bc <u><em>or</em></u><em> </em>A <em>and </em>D.

Explanation:

I just took the quiz and got it correct.

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At the beginning of Section 5.2, it is stated that multiprogramming and multiprocessing present the same problems, with respect
QveST [7]

Answer:

By definition, <u>multiprocessing</u> refers to the processing of multiple processes at the same time by multiple CPUs.

By definition, <u>multiprogramming</u> keeps programs in main memory at the same time and execute them concurrently utilizing a single CPU doing a context switch.

The first difference is that multiprocessing uses multiple CPUs and multiprogramming to utilize context switch to do concurrency in one CPU. Another difference is that multiprocessing is more expensive but more efficient than multiprogramming due that it allows parallel processing.

6 0
3 years ago
What does this say in morse code?
Furkat [3]

Answer:

That it is time for lunch?

Explanation:

4 0
3 years ago
Read 2 more answers
What changes has Sue made so far? Check all that apply.
Alexus [3.1K]

Answer:

Its C, E, and F

Explanation:

On edg

8 0
3 years ago
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A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
The protections from the security software must continue when the device is taken off the network, such as when it is off-grid,
snow_lady [41]

Answer:

third ighjojdudjz

ghddduewjj

wkckdjjj

Explanation:

cchmddkhwmx

ndhdnahdd

jruditndjed

3 0
3 years ago
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