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2 years ago

How many natural numbers lie between thesquares of 48 and 49 *

Mathematics

2 answers:

galben [10]2 years ago

4 0

Answer: 97

Step-by-step explanation:

MA_775_DIABLO [31]2 years ago

3 0

Answer:

97 natural numbers

Step-by-step explanation:

$49^2-48^2\\2401-2304=97$

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Answer:

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Step-by-step explanation:

(6, 4) and (-5, 3)

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Ben bought 4 pairs of jeams and a tshirt that cost 7 he had a cupon for 3 off the price of each pair of jeans amd spent a total

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The dependent variable is the cost of the jeans since it depends on how many jeans Ben bought. An equation that represents the situation is (4j - 12) + 7 = 92

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3 years ago

The number z and 0.8 are additive inverses. Drag and drop 0.8 and z to their correct positions on the number line. Drag and drop

UkoKoshka [18]

we know that

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using a graph tool

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the answer Part a) and part b) in the attached figure

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2 years ago

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notsponge [240]

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3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

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%.

7 0

3 years ago

How many natural numbers lie between thesquares of 48 and 49 *​

How many natural numbers lie between thesquares of 48 and 49 *