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telo118 [61]
4 years ago
13

Suppose that Mr. Warren Buffet and Mr. Zhao Danyang agree to meet at a specified place between 12 pm and 1 pm. Suppose each pers

on arrives between 12 pm and 1 pm at random with uniform probability. What is the distribution function for the length of the time that the first to arrive has to wait for the other?
Mathematics
1 answer:
Natali [406]4 years ago
8 0

Answer:  f(x)=1\ \text{where}\ 12\leq x\leq 13

Step-by-step explanation:

The probability distribution function for uniform distribution in interval [a,b] is given by :_

f(x)=\dfrac{1}{b-a}\ \text{where}\ a\leq x\leq b

Given : Mr. Warren Buffet and Mr. Zhao Danyang agree to meet at a specified place between 12 pm and 1 pm.

Since 1 pm comes after 12 pm, and in 24 hours system we call it as 13 pm.

If each person arrives between 12 pm and 1 pm at random with uniform probability, then the uniform distribution function for this will be :_

f(x)=\dfrac{1}{13-12}=1\ \text{where}\ 1\leq x\leq 12

Hence. the distribution function for the length of the time that the first to arrive has to wait for the other will be :-

f(x)=1\ \text{where}\ 12\leq x\leq 13

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VikaD [51]
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however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
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\\\\\\
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\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
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cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
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% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
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\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
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\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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