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3 years ago

7

engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

pareto charts in this situation. (4 Marks) Please help

1 answer:

lions [1.4K]3 years ago

3 0

The advantage of a pareto chart is to make sure they have all of their tools

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what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .

NISA [10]

Loaded, $H_{Loaded}$ (s) = $\frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL)$ = $\frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S)$ is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

Learn more about Loaded here:

brainly.com/question/20039214

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3 0

1 year ago

Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a

ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time = 5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - $e^{-0.1t$ )³] $e^{-t/MTTF$

so we substitute

R(5) = [ 1 - ( 1 - $e^{-0.1(5)$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 1 - $e^{-0.5$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 0.393469 )³] $e^{-5/MTTF$ = 0.85

[ 1 - 0.06091 ] $e^{-5/MTTF$ = 0.85

0.9391 $e^{-5/MTTF$ = 0.85

$e^{-5/MTTF$ = 0.85 / 0.9391

$e^{-5/MTTF$ = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

7 0

3 years ago

In a pipe of diameter 350mm and length 75mm, water is flowing at a velocity of 2.8m/s. If the Kinematic viscosity of water is 0.

kodGreya [7K]

Explanation:

the answer is in picture

5 0

3 years ago

This assignment will give you more experience on the use of loops In this project, we are going to compute the number of times a

musickatia [10]

Answer:

import java.io.BufferedReader;

import java.io.File;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class Tester {

public static void main(String[] args) throws IOException {

Scanner in=new Scanner(System.in);

System.out.println("Enter a Number =======>");

long N ;

while (!in.hasNextLong()) {

System.out.println("That's not a number!");

in.next();

}

N=in.nextLong();

System.out.println("Number Entered is =======>"+N);

System.out.println("Enter a Digit =======>");

int D;

while (!in.hasNextLong()) {

System.out.println("That's not a number!");

in.next();

}

D=in.nextInt();

System.out.println("Digit Entered is =======> "+ D);

long que=N;

int rem;

int count=0;

while(true){

long temp;

temp = (long)que/10;

rem = (int) (que % 10);

System.out.println(temp+" "+ que+" "+rem);

if(rem==D) count++;

que=temp;

if(que==0) break;

}

System.out.println("The number of "+ D+"'s in "+ N + " is "+ count );;

}

}

Explanation:

Divide the que variable by 10 and assign its result to the temp variable.
Calculate the remainder.
Increment the count if the remainder is equal to the value of D variable and assign the value of que to temp variable.

3 0

3 years ago

CPS 2231: Computer Organization and Programming Programming assignment #1 Concepts: Scanner, loops, input validation, array, met

agasfer [191]

Answer:

import java.util.*;

public class BarChart

{

public static void main(String args[])

{

int arr[]=new int[5];

Scanner sc=new Scanner(System.in);

for(int i=0;i<5;i++)

{

while(true){

System.out.println("Enter today's sale for store "+(i+1)+" (negative value not allowed)");

arr[i]=sc.nextInt();

if(arr[i]>0)

break;

}

}

System.out.println("SALES BAR CHART");

for(int i=0;i<5;i++)

{

System.out.println("Store "+(i+1)+": ");

for(int j=0;j<arr[i];j=j+100)

{

System.out.print("*");

}

System.out.println("");

}

}

}

3 0

3 years ago