153
<u>+121</u>
274
Hope this helps!
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.
L = W + 55
2L + 2W = 4242
L + W = 2121
(W + 55) + W = 2121
2W + 55 = 2121
W = 1033 inches L = 1088 inches
Answer:
Step-by-step explanation:
What did you include in your response? Check all that apply.
There would be an open circle at (2, 1). <u>Yes</u>
There would be a closed circle at (2, 3). <u>Yes</u>
There would be an open circle at (4, 3). <u>Yes</u>
There would be a closed circle at (4, −4). <u>Yes</u>
Endpoints that are not included in the domain of a particular piece of a function are represented by an open circle. <u>Yes</u>