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1 year ago

Write a raptor program that will display the count from 100 to 0 in decrements of 5.

Computers and Technology

1 answer:

S_A_V [24]1 year ago

3 0

A raptor program that will display the count from 100 to 0 in decrements of 5 is shown below.

<h3>What is a raptor program?</h3>

This refers to a programming environment that works on a flowchart basis to help the user to present their algorithms in a visual manner.

Using a raptor program, you can display a count from 100 to 0 in decrements of 5 as shown:

#include <iostream>

#include <string>

using namespace std;

int main()

{

int count;

Count =100;

while (Count<=0)

{

cout << Count << endl;

Count =Count-5;

}

return 0;

}

The result would be a list showing numbers that go from 100 to 0 in decrements of 5.

Find out more on raptor programs at brainly.com/question/15082977

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Explanation:

Given

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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,

$p = 1 - q = 99.98\%$

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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 $Log0.05$

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

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

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

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