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3 years ago

Carla uses the order of operations to solve the equation. 2 + 6 × 3 = b Which is the unknown number?

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Answer:

$b=20$

Step-by-step explanation:

$2 + 6 \times 3 = b$

$2 + 18 = b$

$20= b$

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A 2-yard piece of string costs $5.04. What is the price per foot?

Nikitich [7]

The price per foot would be .84 cents, 1 yard equals 3 feet so 2 yards is 6 feet. if you take $5.04 and divide it by 6 then you get 0.84

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The nile river is 6690 kilometers long.This is 394 kilometers longer than the Amazon River.How long is the Amazon River?

valentinak56 [21]

Answer:

The nile river is 6690 kilometers long.This is 394 kilometers longer than the Amazon River.How long is the Amazon River?

6690-394=6296 kilometers

Step-by-step explanation:

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2 years ago

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Point O is the center of this circle. What is m CAB?<br> A.55<br> B.48<br> C.45<br> D.35

Sophie [7]

Answer is B. 48

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2 years ago

Which linear function represents the line given by the point slope equation y + 7= 2/3(x+6)

Andrews [41]

Hello from MrBillDoesMath!

Answer: f(x) = (2/3)*x -3 -- which is not a provided answer..

Discussion:

We are given that y + 7 = (2/3) * (x + 6). Multiply out the right hand side:

y + 7 = (2/3) * x + (2/3) * 6 = (2/3) *x + 12/3

= (2/3)*x + 4

Subtract 7 from both sides:

y + 7 - 7 = (2/3)*x + 4 - 7

y = (2/3)*x -3

Thank you,

MrB

y + 7= 2/3(x+6)

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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