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Vlada [557]
3 years ago
6

Carla and Rob left a 20% tip after having lunch at a restaurant. The amount of the tip was $6. Carla's lunch cost $15. Write an

equation you can use to find x, the cost of Rob's lunch.
Mathematics
1 answer:
Afina-wow [57]3 years ago
4 0

Robs lunch cost is unknown = x

and we know carla's cost 15  = 15

x + 15 = total lunch cost

now me know 20% is 6

so 20% of total lunch cost is 6

or

(x+15)(.02)  which is 20% of lunch and this is 6

or

(x+15)(.02) = 6  this is the equation to find x

this is the answer, but for kicks lets solve it to to find the cost

.2x + 3 = 6

     - 3    -3  

.2x  = 3

/.2     /.2

x = 15

Robs lunch cost $15 too


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At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
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Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

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As we know the poisson process, we get that

P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

So, for exactly one car would be

P(n=1) is given by

=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

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