Choice J both prices are per a unit
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:
the answer would be D because meadian is what in the middle
So Angle GEN and FED are vertical angles, which means that their angles are congruent. Therefore, Angle FED is 60°
Angle GEN and DEN are a linear pair, which means that they are supplementary angles (add up to 180). To find Angle DEN, we have to form the equation: 60 + DEN = 180
For this, all you have to do is subtract 60 on both sides, and your answer will be: DEN = 120°
Answer:
Step-by-step explanation:
Interest earned during the rest of 2 years and 6 months
40 + 40 + 40 = 120
he will be getting 1000 at the time of maturity ie after 2.5 years .
Total receipt = 1000 + 120 = 1120
investment made = 820
total receipt earned = 1120
profit made = 300
time = 2.5 years
profit = investment x rate of intt x time / 100
300 = 820 x r x 2.5 / 100 where r is rate of interest
r = 30000 / 820 x 2.5
= 14.63 % .