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2 years ago

What are the primary IP addresses for DNS servers hosted on the x128bit, iskytap and cloudparadox domains

Computers and Technology

1 answer:

coldgirl [10]2 years ago

8 0

Answer:

208.91. 197.27 and 50.225.

Explanation:

In the world of computers, the term "IP" stands for Internet Protocol. The IP address or the Internet Protocol address is defined as a numerical label that is provided to every device which is connected to any computer network for communication. It works as providing the location address of the device and also identifies the network interface.

The primary IP addresses for the DNS servers that is hosted on :

-- x128bit is 208.91. 197.27

-- iskytap is 208.91. 197.27

-- cloudparadox is 50.225.

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3 years ago

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

6 0

3 years ago

Five year ago, Amit was three times as old as Arman. Ten years later Amit would be twice as old as Arman. How old is Arman now?

RideAnS [48]

Answer: 50
Explanation: let amit's current age = a and armaan's current age be b.
(a-5) = 3* (b-5)
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10 years later,
(a+10) = 2(b+10)
i.e. a=2b+10 -(ii)
From eqn (i) and (ii),
b=20,
and a=50

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2 years ago

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USPshnik [31]

Answer:

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