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3 years ago

N - 3/5 = 2/5 these two are fractions it says find the value of the variable plz help

Mathematics

1 answer:

lianna [129]3 years ago

4 0

Hello!

To solve this equation, you need to follow SADMEP. SADMEP is subtract, addition, division, multiplication, exponents, and parentheses (if you notice it is PEMDAS backwards) when solving algebraic equations. The goal is to get the variable by itself.

n - 3/5 = 2/5 (add 3/5 to both sides - the denominator of both fractions are equal)

n = 1

Therefore, the value of the variable n is equal to 1.

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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

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

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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natulia [17]

Answer:

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Step-by-step explanation:

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