Answer:
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so
.
What is the probability that a line width is greater than 0.62 micrometer?
That is 
So



Z = 2.4 has a pvalue of 0.99180.
This means that P(X \leq 0.62) = 0.99180.
We also have that


There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Answer : 110 degree
To find angle 1 , we apply outside angle theorem Lets name each point Measurement of arc EF=280 degrees
Measurement of arc GH = 60
Angle D = angle 1
Please refer to the theorem attached below

Now we plug in the values

angle 1 = 110
Measurement of angle 1 = 110 degrees
The way to answer this question is to first make 52% into a decimal:
52% = 0.52
so now we will simply multiply:
84
x.52
once you have multiplied these numbers it comes out to be 43.68.
So your answer is 43.68.
Hope this answer helps! feel free to ask any additional questions :)