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Soloha48 [4]
3 years ago
11

Gavin goes for a run at a constant pace of 9 minutes per mile. Ten minutes later, Lars goes for a run, along the same route, at

a constant pace of 7 minutes per mile. How many minutes does it take for Lars to reach Gavin?
Mathematics
1 answer:
sweet [91]3 years ago
6 0

Answer:  The answer is 35 minutes.


Step-by-step explanation:  Given that Gavin goes for a run at a constant pace of 9 minutes per mile and after 10 minutes, Lars started running along the same route, at a constant pace of 7 minutes per mile. We need to find the number of minutes Lars will take to reach Gavin.

In 9 minutes, distance run by Gavin = 1 mile.

So, in 1 minute, distance travelled by Gavin will be

d_G=\dfrac{1}{9}=\dfrac{7}{63}~\textup{miles}.

Similarly,

In 7 minutes, distance run by Lars = 1 mile.

So, in 1 minute, distance travelled by Lars will be

d_L=\dfrac{1}{7}=\dfrac{9}{63}~\textup{miles}.

Now, since Lars started after 10 minutes, so distance run by Gavin in those 10 minutes will be

d_{G10}=\dfrac{10}{9}=\dfrac{70}{63}~\textup{miles}.

Now, difference between Lars and Gavin's rate of runnings is

\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}.

Therefore, the time taken by Lars to reach Gavin is given by

t=\dfrac{\frac{70}{63}}{\frac{2}{63}}=35.

Thus, the required time is 365 minutes.  


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The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.
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0.52763 is the probability that the time between the next two calls will be 54 seconds or​ less.

0.19285 is the probability that the time between the next two calls will be greater than 118.5 ​seconds.

Step-by-step explanation:

We are given the following information in the question:

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.

The distribution function can be written as:

f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}

The probability for exponential distribution is given as:

P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}

a) P( time between the next two calls will be 54 seconds or​ less)

P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763

0.52763 is the probability that the time between the next two calls will be 54 seconds or​ less.

b) P(time between the next two calls will be greater than 118.5 ​seconds)

p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285

0.19285 is the probability that the time between the next two calls will be greater than 118.5 ​seconds.

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3 years ago
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