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Inessa05 [86]
3 years ago
8

Do these numbers 19.657 < 19.67​

Mathematics
1 answer:
nexus9112 [7]3 years ago
5 0
I dont understand your question.......
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Someone please help me I really need help !!!!
DanielleElmas [232]

Answer:

\frac{71}{100}

<h3>Answer c</h3>

Step-by-step explanation:

\frac{14}{100}  = squash \\  \frac{3}{20}  =  \frac{15}{100}  = tennis \\  \frac{14 +1 5}{100}  =  \frac{29}{100}  \\  \\  \frac{100}{100}  -  \frac{29}{100  }  \\  =  \frac{71}{100} = rugby

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2 years ago
Find the area of the garden?<br> 16 ft<br> 22 ft
Olenka [21]

Answer:

352

Step-by-step explanation:

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3 years ago
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7 0
3 years ago
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.

6 0
3 years ago
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