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3 years ago

If the diameter of a circle is 24, what is the circumference? Round to the nearest thousandth.

Mathematics

2 answers:

kramer3 years ago

7 0

The answer is A.
c=2 x pi x r

Anna35 [415]3 years ago

4 0

A khan academy told me

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Which expression is equivalent to (18)2⋅(19)2

sladkih [1.3K]

The expression that is equivalent to the given expression where the expression is given as (18)2⋅(19)2 is (18 * 19)^2

<h3>How to determine which expression is equivalent to the given expression? </h3>

The expression is given as

(18)2⋅(19)2

Rewrite the above expression properly

So, we have

(18)^2 * (19)^2

The factors in the above expression have the same exponent.

So, the expression can be rewritten as

(18 * 19)^2

Hence, the expression that is equivalent to the given expression where the expression is given as (18)2⋅(19)2 is (18 * 19)^2

Read more about equivalent expression at

brainly.com/question/2972832

#SPJ1

5 0

1 year ago

Y + 2 = -3/4 ( x +1 )

ivann1987 [24]

Answer:

Step-by-step explanation:

3 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

the number of students in the four sixth-grade classs at northside school are 26, 19,34 and 21. Use properties to find the total

Aleks04 [339]

In this case all you do is add all the numbers, answer is 100

4 0

3 years ago

Given the points (8,9) and (2,7) find the slope I need help

Debora [2.8K]

Answer:

1/3

Step-by-step explanation:

7-9=-2

2-8=-6

2/6 is 1/3 when simplified

6 0

3 years ago