Answer:
1/6
Step-by-step explanation:
2/3 divided by 4= 2/3 x 1/4
2/3 x 1/4=2/12
2/12 simplifies to 1/6
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: 0.81
Step-by-step explanation:
=27⋅((3^3)^−1)
Do the brackets first:
3^3 is the same as 3³ so it is <u>27</u>
then:
27^-1 equals <u> 0.03</u>
And lastly:
multiply 27 and 0.03
27(0.03)=<em><u>0.81</u></em>
Answer: y+-1m -1
Step-by-step explanation:
Answer:
Step-by-step explanation:
1.2m = 120 cm
2.16 m = 216cm
120×216 =25920cm
2) 10×18 = 180
25920/180= 144
√144 = 12