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3 years ago

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Mathematics

1 answer:

Damm [24]3 years ago

6 0

Answer:

Unlike fractions: Fractions with different denominators are called, unlike fractions.

In case of two equivalent fractions, the product of the numerator of one fraction and denominator of the second fraction is equal to the product of the denominator of the first fraction and numerator of the second fraction.

A fraction is said to be in form if its numerator and denominator are 5. To add or subtract like fractions, we add or subtract the keeping the same.

Step-by-step explanation:

4 POINT I DONT NO

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0.08

Step-by-step explanation:

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4 years ago

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Bumek [7]

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4 years ago

How does 4/3y = 36??????????

omeli [17]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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 X = number of text messages receive or send in an hour.

The random variable X follows a Poisson distribution with parameter λ.

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