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2 years ago

Solve the equation y – 5x = –7 for y?

Mathematics

1 answer:

eduard2 years ago

5 0

A I believe (sry if I’m wrong)

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Help me PLEASEEE I need help

Nitella [24]

Answer:

area of shaded portion=56 - 1/2 ×3×4=56-6= 50 yd^2

7 0

3 years ago

What is the slope of the line with points (-5, -4) and (2, 7)?

jolli1 [7]

Answer:

m = 11/7

Step-by-step explanation:

m = y2-y1/x2-x1

m = 7-(-4)/2-(-5)

m = 7+4/2+5

m = 11/7

6 0

3 years ago

Please help :):):):):):):):......

Gelneren [198K]

13+13=26
9+9=18
18+26=44

Hope this helps :)

3 0

3 years ago

Read 2 more answers

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

Find the probability of a least one birthday match among a group of 44 people.

Cerrena [4.2K]

Answer:

0.1137= 11.37%

Step-by-step explanation:

Assuming there are 365 days in one year and every people have 1 birthday, then the chance for two people to have the same birthday is 1/365 and the chance they are not is 364/365. We are asked the chance for at least one match among 44 people. The opposite of the condition is that we have 0 matches and easier to calculate. The calculation will be:

P(X>=1)= ~P(X=0) = 1

P(X>=1)=- P(X=0)

P(X>=1)=1 - (364/365)^44

P(X>=1)=1- 0.8862

P(X>=1)=11.37%

3 0

3 years ago