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nexus9112 [7]
3 years ago
15

In the context of web and network privacy, an acceptable use policy is _____.

Computers and Technology
1 answer:
Nikitich [7]3 years ago
4 0
In the context of web and network privacy, an acceptable use policy (AUP) is <span>a set of rules specifying the legal and ethical use of a system and the consequences of noncompliance.
</span><span>These rules restrict the ways in which the network, website or system may be used and sets guidelines as to how it should be used.</span>
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Mental state shift involves what?
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It involves a matter involving doubt, uncertainty, or difficulty that may be solved, problem ... getting into a frame of mind to be creative and solve problems.

Explanation:

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A testing lab wishes to test two experimental brans of outdoor pain long each wiil last befor fading . The testing lab makes six
Simora [160]

Answer:

The answer is "\bold{Brand \ A \ (35, 350, 18.7) \ \ Brand \ B \ (35, 50, 7.07)}"

Explanation:

Calculating the mean for brand A:

\to \bar{X_{A}}=\frac{10+60+50+30+40+20}{6}  =\frac{210}{6}=35

Calculating the Variance for brand A:

\sigma_{A}^{2}=\frac{\left ( 10-35 \right )^{2}+\left ( 60-35 \right )^{2}+\left ( 50-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 20-35 \right )^{2}}{5} \\\\

     =\frac{\left ( -25 \right )^{2}+\left ( 25  \right )^{2}+\left ( 15\right )^{2}+\left ( -5 \right )^{2}+\left ( 5 \right )^{2}+\left ( 15 \right )^{2}}{5} \\\\ =\frac{625+ 625+225+25+25+225}{5} \\\\ =\frac{1750}{50}\\\\=350

Calculating the Standard deviation:

\sigma _{A}=\sqrt{\sigma _{A}^{2}}=18.7

Calculating the Mean for brand B:

\bar{X_{B}}=\frac{35+45+30+35+40+25}{6}=\frac{210}{6}=35

Calculating the Variance for brand B:

\sigma_{B} ^{2}=\frac{\left ( 35-35 \right )^{2}+\left ( 45-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 35-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 25-35 \right )^{2}}{5}

    =\frac{\left ( 0 \right )^{2}+\left ( 10 \right )^{2}+\left ( -5 \right )^{2}+\left (0 \right )^{2}+\left ( 5 \right )^{2}+\left ( -10 \right )^{2}}{5}\\\\=\frac{0+100+25+0+25+100}{5}\\\\=\frac{100+25+25+100}{5}\\\\=\frac{250}{5}\\\\=50

 Calculating the Standard deviation:  

\sigma _{B}=\sqrt{\sigma _{B}^{2}}=7.07

4 0
3 years ago
Protocol 2 - Check for Errors
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Please note that the Problem to be solved from Protocol 1 is not provided hence the general answers. To construct and send, open a network environment a single multi-packet message, simply click "Add Packet" and then click "Send at Once".

<h3>How will the receiver know the order of the packets or if any are missing?</h3>

If the text or message sent does not make any reading sense, or if certain words are jumbled and out of place, then it is clear that something is wrong.

If the messages arrive in a coherent fashion, then the packet was fully received.

<h3>How will the receiver request missed packets and what will the sender do in response?</h3>

Where the users are familiar with the Transmission Control Protocol, lost packets can be detected when there is a timeout. Lost packets are referred to as Dropped packets.

Learn more about Packets at:
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7 0
2 years ago
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