Dagim Mulugeta
Dec 13, 2023
Consider the signal x(t),which consists of single rectangular pulse of unif height, is symmetric about the origin, and has a total width T1.( a) sketch x(t).?
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ChatGPT
Dec 13, 2023
To sketch the signal x(t), we need to consider the given properties:
1. The signal is a single rectangular pulse.
2. The pulse is symmetric about the origin.
3. The pulse has a total width T1.
Let's start by drawing the x-axis and the y-axis.
Next, we draw a rectangular pulse centered at the origin. Since the pulse is symmetric, we can draw half of the pulse on one side of the origin and mirror it on the other side.
The width of the pulse is T1, so we draw the pulse from -T1/2 to T1/2 on the x-axis.
The height of the pulse is uniform, so we draw the pulse with a constant value on the y-axis.
The resulting sketch of x(t) would look like a rectangular pulse centered at the origin, with a total width of T1 and a uniform height.
1. The signal is a single rectangular pulse.
2. The pulse is symmetric about the origin.
3. The pulse has a total width T1.
Let's start by drawing the x-axis and the y-axis.
Next, we draw a rectangular pulse centered at the origin. Since the pulse is symmetric, we can draw half of the pulse on one side of the origin and mirror it on the other side.
The width of the pulse is T1, so we draw the pulse from -T1/2 to T1/2 on the x-axis.
The height of the pulse is uniform, so we draw the pulse with a constant value on the y-axis.
The resulting sketch of x(t) would look like a rectangular pulse centered at the origin, with a total width of T1 and a uniform height.
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Kenenisa Waktole
Dec 13, 2023
To find the probability for the given scenarios, we can use the Poisson probability formula:
P(x;λ) = (e^(-λ) * λ^x) / x!
Where
λ = 12 (average number of incidents per year)
x = the number of occurrences for each scenario
a) Probability that exactly three will experience anaphylaxis:
P(3;12) = (e^(-12) * 12^3) / 3!
P(3;12) ≈ 0.1114 or 11.14%
b) Probability that at least two will experience anaphylaxis:
P(at least 2;12) = 1 - P(0;12) - P(1;12)
P(at least 2;12) = 1 - (e^(-12) * 12^0) / 0! - (e^(-12) * 12^1) / 1!
P(at least 2;12) ≈ 0.8647 or 86.47%
c) Probability that at most two will experience anaphylaxis:
P(at most 2;12) = P(0;12) + P(1;12) + P(2;12)
P(at most 2;12) = (e^(-12) * 12^0) / 0! + (e^(-12) * 12^1) / 1! + (e^(-12) * 12^2) / 2!
P(at most 2;12) ≈ 0.2642 or 26.42%
P(x;λ) = (e^(-λ) * λ^x) / x!
Where
λ = 12 (average number of incidents per year)
x = the number of occurrences for each scenario
a) Probability that exactly three will experience anaphylaxis:
P(3;12) = (e^(-12) * 12^3) / 3!
P(3;12) ≈ 0.1114 or 11.14%
b) Probability that at least two will experience anaphylaxis:
P(at least 2;12) = 1 - P(0;12) - P(1;12)
P(at least 2;12) = 1 - (e^(-12) * 12^0) / 0! - (e^(-12) * 12^1) / 1!
P(at least 2;12) ≈ 0.8647 or 86.47%
c) Probability that at most two will experience anaphylaxis:
P(at most 2;12) = P(0;12) + P(1;12) + P(2;12)
P(at most 2;12) = (e^(-12) * 12^0) / 0! + (e^(-12) * 12^1) / 1! + (e^(-12) * 12^2) / 2!
P(at most 2;12) ≈ 0.2642 or 26.42%
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Kenenisa Waktole
Dec 13, 2023
In a certain population an average of 13 new cases of esophageal cancer are diagnosed each year. If the annual incidence of esophageal cancer follows a Poisson distribution, find the probability that in a given year the number of newly diagnosed cases of esophageal cancer will be:A. Exactly 10 cases B. At least three cases C. No more than 3D. Between nine and 12, inclusiveE. Fewer than two
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Kenenisa Waktole
Dec 13, 2023
In a certain population an average of 13 new cases of esophageal cancer are diagnosed each year. If the annual incidence of esophageal cancer follows a Poisson distribution, find the probability that in a given year the number of newly diagnosed cases of esophageal cancer will be:A. Exactly 10 cases B. At least three cases C. No more than 3D. Between nine and 12, inclusiveE. Fewer than two
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