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Aleks [24]
3 years ago
13

Can someone please help me?!

Mathematics
1 answer:
____ [38]3 years ago
7 0

Answer:

The answer is 5^10 or 9765625

Step-by-step explanation:

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Answer:

It is A: 11 to 1 :)

It's true!

(edited my answer btw, you have to simply also welp)

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Solve the equation <br> 13(x + 1) + 2x = 2
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businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

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Tukar 88 peratus kepada pecahan yang termudah​
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kya hai mujhye kuchh samajh na aaraha

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If you added more context then I would probably answer your question
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