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3 years ago

Help, help help. just give answers

Mathematics

2 answers:

Fynjy0 [20]3 years ago

5 0

To solve this problem, there are a couple things we need to do. First of all, we need to figure out how much more she's running every day. On day 1, she ran 3/16 of a mile. On day 2, she ran 3/8 of a mile. On day 3, she ran 9/16 of a mile.

How much are we adding each time? To find that out, we need all of our fractions to have the same denominator so we can compare. Let's give 3/8 a denominator of 16. 8*2=16. What you do on one side of the fraction bar, you MUST do to the other: 3*2=6. So, 3/8 = 6/16.

Day 1: 3/16. Day 2: 6/16. Day 3: 9/16. Now do you see the pattern? Every day, she runs 3/16 of a mile more. That's the answer to part 1: 3/16.

So, how long will it take her to get over 1 mile? Let's continue. Day 3: 9/16. Day 4: 9+3=12, so that's 12/16. Day 5: 15/16. Day 6: 18/16! Since the numerator is bigger than the denominator, guess what? That's more than one mile!! So on the 6th time she runs, she ran 18/16 miles. Keep in mind that 18/16 = 9/8 = 1 1/8 miles.

Answers: 3/16; 6th; 1 1/8 (or 9/8)

ohaa [14]3 years ago

4 0

The answers to your questions are: 1st question:3/16 2nd question:6th 3rd question:11/8.
Hope this helps! (:

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Step-by-step explanation:

Since we have given that

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

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P(n=1) is given by

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

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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,

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

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