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pogonyaev
4 years ago
9

What is the circumference of the circle in terms of / pi? The radius is 30 in.

Mathematics
2 answers:
Alinara [238K]4 years ago
5 0
Circumference formula is C= 2 * pi * r C=188.5 (pi 60)
Ksenya-84 [330]4 years ago
4 0

Answer:

Circumference of circle in terms of π is 60\pi\:inches

Step-by-step explanation:

Given: Radius of the circle = 30 inches.

To find : Circumference of the circle.

When r is the radius of the circle, then circumference is equal to 2\pi r.

Here, radius , r = 30 inches.

So, Circumference = 2\pi r=2\times\pi\times30=60\pi

Therefore, Circumference of circle in terms of π is 60\pi\:inches

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Answer: 12/25 is 0.48

because 12 divided into 25 equals 0.48

Step-by-step explanation:I’m sorry if I answered it too late

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3 years ago
I need to find the equation of the line
deff fn [24]

Answer:

y = 3x + 3

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

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m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 1, 0) and (x₂, y₂ ) = (0, 3) ← 2 points on the line

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3 0
2 years ago
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Given quadrilateral ABCD is congruent to Quadrilateral A'B'C'D', What is the measure of B'C'?
Vadim26 [7]
The measure of B’C’ is 18 inches
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3 years ago
You are purchasing 2 shirts at $15.00 each and 1 pair of jeans at $30.00 each. You have a coupon for 10% off. How much is the to
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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

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