The probability that the error occurred when Engineer 2 made the mistake is 0.462 on the other hand the probability that the error occurred when the engineer 1 made the mistake is 0.538
Explanation:
Let 
denote the event that the 1st engineer does the work.so we write
=0.7
Let 
denote the event that the 2nd engineer does the work .So we write
=0.3
Let O denote the event during which the error occurred .so we write
=0.02(GIVEN)
=0.04(GIVEN)
- The probability that the error occurred when the first engineer performed the work is
- The probability that the error occurred when the first engineer performed the work is
Now we need to find when did the error in the work occur so we will compare the probability of the work done by <u>engineer 1 </u>and <u>engineer 2 </u>
<u></u>
<u>lets find the Probability of the Engineer 1</u>
<u>Using Bayes theorem,we get</u>
<u></u>
<u></u> =0.02*0.7/0.02*0.7+0.04*0.3 = 0.014/0.026=0.538
<u>lets find the Probability of the Engineer 2</u>
<u></u> =0.04*0.3/0.02*0.7+0.04*0.3=0.012/0.026=0.462
Since ,0.462<0.538 so it is more prominent that the Engineer 1 did the work when the error occurred
Answer:
Fatigue occurs under conditions of high elastic stress, high stress fluctuations and high rate of loading
Explanation:
According to many definition of fatigue failure the fatigue occurs when in an especifyc point of the object there is involved many forces and tensions.
That tensions needs to be big in magnitud, de variations of the efforts it has to be with a lot of amplitude and the loading in the object it has to be with a lot of number of cycles.
If in the all of these three conditions are present the fatigue failure it would appear.
Answer:
//Program was implemented using C++ Programming Language
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
unsigned int second_a(unsigned int n)
{
int r,sum=0,temp;
int first;
for(int i= 1; I<=n; i++)
{
first = n;
//Check if first digit is 3
// Remove last digit from number till only one digit is left
while(first >= 10)
{
first = first / 10;
}
if(first == 3) // if first digit is 3
{
//Check if n is palindrome
temp=n; // save the value of n in a temporary Variable
while(n>0)
{
r=n%10; //getting remainder
sum=(sum*10)+r;
n=n/10;
}
if(temp==sum)
cout<<n<<" is a palindrome";
else
cout<<n<<" is not a palindrome";
}
}
}
Explanation:
The above code segments is a functional program that checks if a number that starts with digit 3 is Palindromic or not.
The program was coded using C++ programming language.
The main method of the program is omitted.
Comments were used for explanatory purpose.
