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
![P(E_{2} /O)](https://tex.z-dn.net/?f=P%28E_%7B2%7D%20%2FO%29)
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:
Amount of fuel used per year is supposed to be 34150 KJ/kg
STP stands for standard temperature pressure and NTP stands for normal temperature pressure
Answer:
135 hour
Explanation:
It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.
We have to find the time necessary to achieve the same concentration at a 6 mm position.
we know that
where x is distance and t is time .As the temperature is constant so D will be also constant
So
then
we have given
and we have to find
putting all these value in equation
![\frac{2^2}{15}=\frac{6^2}{t_2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5E2%7D%7B15%7D%3D%5Cfrac%7B6%5E2%7D%7Bt_2%7D)
so
Answer:
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