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Anvisha [2.4K]
2 years ago
12

Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis

tribution, with a mean of 30,000 miles. Find the probability that one of these tires, bought today, will last the following number of miles:a.Over 30,000 milesb.Over 30,000 miles, given that it already has gone 15,000 miles.
Mathematics
1 answer:
wariber [46]2 years ago
7 0

Answer:

a) P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

b) P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}, x>0

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

X \sim Exp(\lambda=\frac{1}{30000})

The cumulative distribution function is given by:

F(X) = 1- e^{-\frac{x}{\mu}}

Part a

We want to find this probability:

P(X>30000) and for this case we can use the cumulative distribution function to find it like this:

P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

Part b

For this case w want to find this probability

P(X>30000|X>15000)

We have an important property on the exponential distribution called "Memoryless" property and says this:

P(X>a+t| X>t)=P(X>a)  

On this case if we use this property we have this:P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)

We can use the definition of the density function and find this probability:

P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

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CaHeK987 [17]
Let's say "p" people were going to the expedition initially, and the cost for each was "c", now, we know the total cost is 1800, so for "p",  folks that'd be  1800/p how much each one cost, namely, how many times "p" goes into 1800.

well, prior to leaving, 15 dropped out, so that leaves us with " p - 15 ", and the cost "c" bumped up to " c + 27 " for each.

\bf \begin{cases}
\cfrac{1800}{p}=\boxed{c}\\\\
\cfrac{1800}{p-15}=c+27\\\\
----------\\
\cfrac{1800}{p-15}=\boxed{\cfrac{1800}{p}}+27
\end{cases}\\\\
-------------------------------\\\\
\cfrac{1800}{p-15}={\cfrac{1800}{p}}+27\impliedby 
\begin{array}{llll}
\textit{let's multiply both sides by}\quad  p(p-15)\\
\textit{to get rid of the denominators}
\end{array}

\bf 1800p=1800(p-15)+27[p(p-15)]
\\\\\\
1800p=1800p-27000+27(p^2-15p)
\\\\\\
0=-27000+27(p^2-15p)\implies 0=-27000+27p^2-405p
\\\\\\
\textit{now, let's take a common factor of }27
\\\\\\
0=p^2-15p-1000\implies 0=(p-40)(p+25)\implies p=
\begin{cases}
\boxed{40}\\
-25
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well, you can't have a negative value of people... so it has to be 40.

so, 40 folks were initially going, then 15 dropped out, how many went on the expedition?  40 - 15.
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You have 20 quarters. You find 40% more in your room. Then you go to the store and spend 50% of them. How much money do you have
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Answer:

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Step-by-step explanation:

First fond 40% of 20

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28×50/100 =14

So now you are left with 28-14 =14

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Answer:

The answer to number 1 is A.

Step-by-step explanation:

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Answer:

46

Step-by-step explanation:

94-48

46

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