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White raven [17]
4 years ago
12

Is the simplified form of 2√3 - 2√3 rational?

Mathematics
1 answer:
Mila [183]4 years ago
5 0
Yes, the answer is rational.
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2 years ago
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krok68 [10]

Answer:

The hypotenuse of a right triangle is 4m longer than the shorter leg and 2m longer than the longer leg. What are the lengths of the sides?

Just for variety, consider the hypotenuse = h, the short leg h-4 and the long leg h-2.

c^2 = a^2 + b^2; so h^2 = (h-4)^2 + (h-2)^2; h^2 = h^2 - 8h + 16 +h^2 -4h +4;

h^2 -12h +20 = 0 factors to (h - 10)(h - 2) = 0 so h = 2 or h = 10

Since (h - 2) = (2 - 2) = 0 and a triangle cannot have a side of zero length, 10 is the length of the hypotenuse.

h^2 = (h-4)^2 + (h-2); 10^2 = (10-4)^2 + (10-2); (10)^2 = (6)^2 + (8)

The hypotenuse is 10 cm, short leg 6 cm and long leg 8 cm.

5 0
3 years ago
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

brainly.com/question/24756209

#SPJ4

7 0
2 years ago
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