All categories

3 years ago

Lsplslsplpsplslpsllspslpslpslpslpsplspslpslpslpslpslsplsplspls

Mathematics

2 answers:

nata0808 [166]3 years ago

8 0

Answer: 257.61

Step-by-step explanation:

The formula of a circle for circumference is C = 2 π r

When you follow that formula correctly, all your answers will be correct.

NOTE: This formula does not refer to the area or diameter.

Ivan3 years ago

6 0

Answer:

257.61

Step-by-step explanation:

FORMULA= C=2πr=2·π·41≈257.6106

there is a screenshot included too

You might be interested in

F x and r are inversely proportional and x = 9 when r = 72, then what is x when r = 40?

Oksi-84 [34.3K]

Answer:

16.2

Step-by-step explanation:

r = k/x

when x = 9 and r = 72 ,

72 = k/9

k = 72*9

When r = 40 ,

x = k/r

x = 72*9/40 = 16.2

8 0

3 years ago

Voting rates in presidential elections from 1996-2012 are modeled below

Ray Of Light [21]

Juiuiiijkkkkkkkkkkkkkkkkkk

3 0

3 years ago

The computer that controls a bank's automatic teller machine crashes a mean of 0.60.6 times per day. What is the probability tha

OLEGan [10]

Answer:

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 per day.

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash more than 3 times?

Either it crashes 3 or less times, or it crashes more than 3 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 3) + P(X > 3) = 1$

We want P(X > 3). So

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X = 3) = \frac{e^{-4.2}*(4.2)^{3}}{(3)!} = 0.1852$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0150 + 0.0630 + 0.1323 + 0.1852 = 0.3955$

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.3955 = 0.6045$

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.

3 0

4 years ago

5/8 - 1/2 =<br><br> 1/8<br> 2/3<br> 1/2

BartSMP [9]

Answer:

5/8 - 1/2 = 1/8

4 0

3 years ago

TC= 0.0001Q^2-0.02Q+5+500Q

Leona [35]

Answer: 0

−

Step-by-step explanation:

5 0

3 years ago