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.
Where an, an-1,a2, a1, a0 are constants. We call the term containing the highest power of x the leading term, and we call an the leading coefficient. The degree of the polynomial is the power of x in the leading term. We have already seen degree 0, 1, and 2 polynomials which were the constant, linear, and quadratic functions, respectively. Degree 3, 4, and 5
D=12
Because if you write down the equation 4x-8=40 and plug in 12. 4 times 12 is 48
. 48-8 equals 40
X + x - 5 + 3x + 25 = 180
5x + 20 = 180
5x = 160
x = 32
m<F = x - 5
m<F = 32 - 5
m<F = 27
Well, this is a Pythagorean theorem problem. A^2 + B^2 = C^2, where C^2 is equal to the hypotenuse. 8 squared equals 64, so c^2 = 64 cm. The other leg can be represented by A^2, which is 36 cm. 36 + ? = 64. 64 - 36 = 28, so B^2 equals 28. Now, to find the measurement of the other leg, we need the square root of 28. The square root of 28 is 5.3 cm.
Your final answer is 5.3 cm.
