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3 years ago

Use unit analysis to determine the unit of measurement for the expression 10x + 25y

Computers and Technology

1 answer:

zimovet [89]3 years ago

8 0

Answer:

Given 10x + 10y

It has to be ensured that x, and y are a part of the same unit system, like if y is meter, then x can be kilometer, centimeter, or milimeter. And we will multiple and divide the bigger or smaller out of the above to get them to the unit, we want them to be in. Like if y is in meter, and x is given in centimeter then, we should divide x by 100 to convert it to meter, and then we can add both.

x=3 centimeter, and y= 2 meters

x = .03 meters

hence = 0.03 *10 + 10*2 =20.3 meters

Explanation:

Please check the answer section.

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Write a short program that allows the user to input a positive integer and then

pogonyaev

Answer:

// code in C++.

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variables

int n;

cout<<"Enter a positive number:";

// read number

cin>>n;

// check number is positive or not

while(n<0)

{

// if number is negative

cout<<"Wrong input!!"<<endl;

// ask again to enter again

cout<<"Enter again:";

// read number again

cin>>n;

}

// generate random number between 1 to n

int ran=rand()%n +1;

// print random number

cout<<"Random number between 1 to "<<n<<" is: "<<ran<<endl;

return 0;

}

Explanation:

Read a number from user.Then if input number is negative then ask user to enter a positive number again.After this generate a random number between 1 to n.Print that random number.

Output:

Enter a positive number:-12

Wrong input!!

Enter again:9

Random number between 1 to 9 is: 2

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3 years ago

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Pavlova-9 [17]

Answer:

javascript, python, c#

Explanation:

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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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B i'm sure of it. 99% sure

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3 years ago

Use unit analysis to determine the unit of measurement for the expression 10x + 25y​

Use unit analysis to determine the unit of measurement for the expression 10x + 25y