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2 years ago

What is the slope of the line whose equation is y - 2/3 = 0?

Mathematics

1 answer:

elena55 [62]2 years ago

6 0

Answer:

The slope is 0.

Step-by-step explanation:

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TELE

Aleonysh [2.5K]

Answer:

$23271.49

Step-by-step explanation:

The function $C(t) = C(1 + r)^{t}$ models the rise in the cost of a product that has a cost of C today, subject to an average yearly inflation rate of r for t years.

Now, if the average annual rate of inflation over the next 8 years is assumed to be 2.5% then we have to find the inflation-adjusted cost of a $19100 motorcycle after 8 years.

Therefore, the cost will be $C(8) = 19100(1 + \frac{2.5}{100} )^{8} = 23271.49$ dollars. (Answer)

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2 years ago

elliot drove 202.8 miles and used 6.5 gallons of gasoline.How many miles did he travel per gallon of gasoline?

Scrat [10]

I think the answer is 31.2 miles per gallon

5 0

3 years ago

Read 2 more answers

You measured the width of your front door to be 3 feet and 2 inches. the chair you are trying to get through the door measured t

Citrus2011 [14]

18 inches or 1 foot and 6 inches
multiply 3 feet by 12 and add two to get the width of the door in inches then subtract 20 inches from it to get the difference

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2 years ago

Read 2 more answers

1. Georgia did an experiment. She started out with 700 bacteria cells. She found that the growth rate of the bacteria cells was

Kitty [74]

panocha

Step-by-step explanation:

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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