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TiliK225 [7]
3 years ago
5

Jasmine wants electrical current to be able to pass through a certain switch on a circuit board that she is designing. What stat

e should that switch be?
state 1

state 0

input

output
Computers and Technology
1 answer:
Alenkinab [10]3 years ago
3 0
State 1 I don’t know if it’s right tho
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Using a browser would help in doing that.
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The fundamental difference between a switch and a router is that a switch belongs only to its local network and a router belongs
Alla [95]

Answer:

The answer is True

Explanation:

Switches are responsible for connecting computers within a network and it belongs only to its local network. Its responsibility is to filter and forward packets between LAN segments. They operate on layer two (2) and sometime layer three (3) of the OSI Model.

Routers on the other hand, are responsible for the interconnections of two or more networks. They forward data packets between networks. They reside at gateways, exactly where two or more networks connects. And they use communication protocols to effectively and efficiently communicate among two or more host.

7 0
3 years ago
What is occurring when an attacker manipulates commonplace actions that are routinely performed in a business?
serg [7]

Answer:

Vulnerable business processes, also called business process compromise (BPC), occurs when an attacker manipulates commonplace actions that are routinely performed.

4 0
3 years ago
TWO QUICK QUESTIONS
andrezito [222]
I'm guessing 8? But I'm not 100% positive 
5 0
4 years ago
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
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