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3 years ago

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A programmer is writing a program that is intended to be able to process large amounts of data. Which of the following considera

tions is LEAST likely to affect the ability of the program to process larger data sets?A. How long the program takes to run
B. How many programming statements the program contains
C. How much memory the program requires as it runs
D. How much storage space the program requires as it runs

Computers and Technology

1 answer:

Dmitry [639]3 years ago

8 0

Answer:

B. How many programming statements the program contains

Explanation:

If a program is intended to be able to process large amounts of data, then the ability of the program to process larger data sets is heavily affected by

execution time
size of memory the program requires
storage space the program requires as it runs.

The length of the statements the program contains does not much affect program performance.

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Answer:

The smallest number of servers required is 3 servers

<em />

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;

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$p + q = 1$ '

Subtract p from both sides

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Substitute $p = 95\%\\$

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<em>To get an expression for one server</em>

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 $q = 0.05$

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Take Log of both sides

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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 $Log0.05$

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

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

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

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

$n = 2.84310893421$

$n = 3 (Approximated)$

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Answer:

The answer to this question is given below in the explanation section.

Explanation:

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