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2 years ago

Which command could you use to change to the /usr directory using a relative pathname?

Computers and Technology

1 answer:

swat322 years ago

6 0

Answer:

To change directories, use the cd command. This command by itself will always return you to your home directory; moving to any other directory requires a pathname. You can use absolute or relative pathnames.

Explanation:

I hope it's help

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Jenny wants to create a résumé after a two-years gap. What should she consider?

gavmur [86]

She should consider jail time.

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3 years ago

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Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent server

sattari [20]

Answer:

The smallest number of servers required is 3 servers

Explanation:

Given

$Reliability = 99.98\%$

$Individual Servers = 95\%$

Required

Minimum number of servers needed

Let p represent the probability that a server is reliable and the probability that it wont be reliable be represented with q

Such that

$p = 95\%\\$

It should be noted that probabilities always add up to 1;

So,

$p + q = 1$ '

Subtract p from both sides

$p - p + q = 1 - p$

$q = 1 - p$

Substitute $p = 95\%\\$

$q = 1 - 95\%$

Convert % to fraction

$q = 1 - \frac{95}{100}$

Convert fraction to decimal

$q = 1 - 0.95$

$q = 0.05$

------------------------------------------------------------------------------------------------

To get an expression for one server

The probabilities of 1 servers having 99.98% reliability is as follows;

$p = 99.98\%$

Recall that probabilities always add up to 1;

So,

$p + q = 1$

Subtract q from both sides

$p + q - q = 1 - q$

$p = 1 - q$

So,

$p = 1 - q = 99.98\%$

$1 - q = 99.98\%$

Let the number of servers be represented with n

The above expression becomes

$1 - q^n = 99.98\%$

Convert percent to fraction

$1 - q^n = \frac{9998}{10000}$

Convert fraction to decimal

$1 - q^n = 0.9998$

Add $q^n$ to both sides

$1 - q^n + q^n= 0.9998 + q^n$

$1 = 0.9998 + q^n$

Subtract 0.9998 from both sides

$1 - 0.9998 = 0.9998 - 0.9998 + q^n$

$1 - 0.9998 = q^n$

$0.0002 = q^n$

Recall that $q = 0.05$

So, the expression becomes

$0.0002 = 0.05^n$

Take Log of both sides

$Log(0.0002) = Log(0.05^n)$

From laws of logarithm $Loga^b = bLoga$

So,

$Log(0.0002) = Log(0.05^n)$ becomes

$Log(0.0002) = nLog(0.05)$

Divide both sides by $Log0.05$

$\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}$

$\frac{Log(0.0002)}{Log(0.05)} = n$

$n = \frac{Log(0.0002)}{Log(0.05)}$

$n = \frac{-3.69897000434}{-1.30102999566}$

$n = 2.84310893421$

$n = 3 (Approximated)$

Hence, the smallest number of servers required is 3 servers

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3 years ago

which process consists of one application program following a logical access path (lap) and then a dbms following a physical acc

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A DBMS acts as a conduit between the database and the applications that will use it, enabling users to access, modify, and control how the data is arranged and optimized.

Data points that are connected to one another are stored and accessible in a relational database, which is a form of database. The relational model, an easy-to-understand method of representing data in tables, is the foundation of relational databases. Data items are requested from the database by application programs, which are created using a combination of the DBMS's data manipulation language and a traditional programming language. The DBMS locates and provides the data elements requested by the application programs.

Learn more about database here-

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1 year ago

Suppose an IP packet is fragmented into 10 fragments, each with a 1% (independent) probability of loss. To a reasonable approxim

Marysya12 [62]

Answer:

a. 0.01

b. 0.001

c. The identification field of the packet fragment can be used to uniquely identify and collate the fragments lost in transmission.

Explanation:

The probability of losing a packet is 10% or 0.1, so the probability of losing the packet twice during transmission;

= 0.1 x 0.1 = 0.01

When any fragments have been part of the transmission, the probability of the packet is dependent on the fragments;

= 0.01 x 0.1 = 0.001

The identification field is a unique 16-bit value assigned to an IPv4 packet, when a packet is fragmented for transmission, its field is used to collate the unique fragments in the packet.

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2 years ago

What is word processors

otez555 [7]

A word processor is a computer program or device that provides for input, editing, formatting and output of text, often with additional features.

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