The line width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micromete
r and a standard deviation of 0.05 micrometer.
What is the probability that a line width is greater than 0.62 micrometer?
What is the probability that a line width is between 0.47 and 0.63 micrometer?
The line width of 90% of samples is below what value?
What is the probability that a line width is greater than 0.62 micrometer?
What is the probability that a line width is between 0.47 and 0.63 micrometer?
The line width of 90% of samples is below what value?
1 answer:
Answer:
a. 0.82%
b. 71.11%
c. 0.564 micrometer
Step-by-step explanation:
To find the z value for each measurement, we must calculate and determine the percentage they correspond to, and the difference will give us the percentage between those two statistics.
The equation for z is:
z = (x - m) / (sd)
Here, x is the value being evaluated, m denotes the mean, and sd represents the standard deviation.
a.
For 0.62 copies, we calculate:
z = (0.62 - 0.5) / (0.05)
z = 2.4
This translates to 0.9918.
Therefore, p (x > 0.62) = 1 - 0.9918
p (x > 0.62) = 0.0082 = 0.82%
b.
For 0.47 copies, we have:
z = (0.47 - 0.5) / (0.05)
z = -0.6, which equates to 0.2742.
For 0.63 copies:
z = (0.63 - 0.5) / (0.05)
z = -2.6, which yields 0.9953.
Hence, p (0.47 > x > 0.63) = 0.9953 - 0.2742
p (0.47 > x > 0.63) = 0.7211 = 72.11 %
c.
For x copies, we find:
p = 0.9 corresponds to z = 1.28.
Thus, 1.28 = (x - 0.5) / (0.05)
From which we derive:
x = 1.28*0.05 + 0.5
x = 0.564 micrometer
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Question:
Which statement is accurate concerning the function’s discontinuities
A) There are gaps at x = 7 and.
B) Asymptotes exist at x = 7 and.
C) Asymptotes exist at x = –7 and.
D) Gaps are present at (–7, 0) and.
Answer:
B) Asymptotes exist at x = 7 and
Step-by-step explanation:
Given:
Goal:
Identify the correct statement
We need to factor the denominator first.
To express x in terms of (3x+4) and (x-7):
3x + 4 =
3x = -4
Divide both sides by 3:
x - 7
x = 7
Next, evaluate the limit when and at (x = 7)
lim f(x) as = ±∞
lim f(x) as (x=7) = ±∞
Since both scenarios result in the denominator approaching zero, they represent asymptotes.
Thus, asymptotes are found at and x=7
Option B is determined to be correct
Answer:
Approximately 59 stacked cups.
Step-by-step explanation:
We have the following measurements:
Height of one cup = 12.5 cm,
Height of two cups stacked = 14 cm,
Height of three cups stacked = 15.5 cm,
...and so on.
This situation can be described by an arithmetic sequence,
12.5, 14, 15.5,....
The first term is defined as a = 12.5,
with a common difference of d = 1.5 cm.
Thus, the height of x stacked cups is given by
As per the problem,
h(x) = 200
⇒ 1.5x + 11 = 200
⇒ 1.5x = 189
⇒ x = 59.3333333333 ≈ 59.
Therefore, you will need approximately 59 stacked cups.