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 principal amount was $23,393.45
Step-by-step explanation:
The total amount paid on a 35 year loan was $98,000 at the rate of interest 4.1%
We will calculate Principal amount by this formula

Where A = amount (98,000)
P = Principal amount (P)
r = rate of interest 4.1% (0.041)
n = number of compounding interest monthly (12)
t = time (35 years)



98,000 = P(4.189386)
= 4.189386P = 98,000
P = 
P = 23,392.4494 ≈ $23,392.45
The principal amount was $23,393.45
The answer is 74.32 y ok it welcome
Let D be dogs and C be cats
<em>dogs, d, is initially five less than twice the number of cats, c</em>
D + 5 = 2C
<em>If she decides to add three more of each, the ratio of cats to dogs will be</em>
D + 8 = 2C + 3
<em>Could Bea's Pet Shop initially have 15 cats and 20 dogs?</em>
Simply plug in the numbers
20 + 5 = 2(15)
This is clearly not true: 25 does not equal 30
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Answer:
d =7.6
Step-by-step explanation:
