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:
Possible lengths of the silk fibre 'x' are 100, 75, 60, 50 and 30 yards respectively.
Step-by-step explanation:
Let the length of the silk fiber = x yards.
Let the number of silk fibres = a.
It is given that the silk fibers between 3 to 10 yards are combined to form a silk thread of 300 yards.
So, we get the equation,
, where 
i.e.
, 
So, the possible values of 'a' are 3, 4, 5, 6 an 10.
Thus, the possible lengths of the silk fibre 'x' are 100, 75, 60, 50 and 30 yards respectively.
4.5 litres of fruit punch with 0.5 litres of water.
13.5 /4.5 = 3. Multiply 0.5 by 3 to get 1.5 litres of water. Therefore, 13.5 litres of fruit punch concentrate needs 1.5 litres of water. 13.5 + 1.5 = 15 litres of fruit punch in total.
Answer:
wow
Step-by-step explanation:
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