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IgorC [24]
2 years ago
5

Please answer quick!!

Mathematics
1 answer:
rewona [7]2 years ago
7 0

Answer: B

Step-by-step explanation:

\frac{2}{7k}(k-7)

\frac{2k-14}{7k}

\frac{2k}{7k}-\frac{14}{7k}

\frac{2}{7}-\frac{2}{k}

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Which statement about the quadratic equation below is true? x2 + 4x + 1 = 0.
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5 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
2 years ago
Stacy rolls a pair of six-sided fair dice.
Nutka1998 [239]

Answer:

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number is 24/36 = 2/3

The two events are mutually exclusive

Step-by-step explanation:

we first need to create a table of the sample space from the experiment, the sum of the numbers rolled. Find the attached for a depiction of the possible outcomes.

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number will be given by counting the numbers that are either even or multiples of 3 and then dividing by the number of possible outcomes, 36.

In our case this will be;

24/36 = 2/3

5 0
3 years ago
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