If f (x) = 3 [X-2] what is f(4.7) answer

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

7 0

Answer:

Step-by-step explanation:

[x-2]=[4.7-2]=[2.7]=2

f(4.7)=3*2=6

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Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

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7 0

2 years ago

What is the answer to 64.521-1.400

goldenfox [79]

Subtract 1.400 from 64.521 and you get 63.121.

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3 years ago

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