All categories

3 years ago

What quantity will correctly complete the formula? arc length = ?/360circumference

Mathematics

1 answer:

jeka57 [31]3 years ago

8 0

? should be the angle measure of the arc.

You might be interested in

A bicycle wheel has a 10-inch radius.How far does the wheel travel in one full turn

wlad13 [49]

Answer:

Approximately 62.83 inches.

Exactly 20π inches.

Step-by-step explanation:

It's simply the wheel's circumference.

C = 2πr = 2(10)(π) = 20π ≈ 62.83 inches

4 0

3 years ago

Read 2 more answers

Which problem is solved using this model? 15÷15 3÷115 3÷15 15÷15 https://static.k12.com/nextgen_media/assets/1503272-MAT

Pie

Answer:

As the model is shown below.

As per model a thing is divided into five parts.There are three things and each one of them is being divided into five equal parts.

The value of fifth part of each of them is 1/5.

If we consider each part of a thing as one,

So, total number of parts = 15

Total thing=3

So, the answer of the above model is 3÷ 15.

Out of all the options which are 15÷15 , 3÷115 , 3÷15, 15÷15 →3÷15 is correct.

5 0

3 years ago

Need help with all of these!!! ASAP!! 15, 16, 17, 18!!!

vodomira [7]

Answer:

15. A

16. B

17. B

18. C

Step-by-step explanation:

8 0

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

Use PEMDAS to solve this problem 6÷2(1+2)=?

Finger [1]

The answer would be 8. hope this helps

8 0

3 years ago

Read 2 more answers