None of the answers are correct because it’s definitely not “Given” because x=20 is like an answer. It’s also not “Distributive Property”, “subtraction Property”, and or “Additional Property” b/c there’s nothing that can be distributed, subtract or add. It’s just an answer.

3 0

3 years ago

Read 2 more answers

Diane keeps her ribbons in a drawer. She has 3 white ribbons, 6 pink ribbons, 2 yellow ribbons, and 7 blue ribbons. Without look

Akimi4 [234]

Answer:

1/3

Step-by-step explanation:

She has 18 ribbons in total and she has 6 pink ribbons so she has a 6/18 chance or 1/3 chance of pulling a pink ribbon

6 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 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

identify the transformations that map rectangle A to rectangle A'. A) reflect over y-axis, reflect over line y = 5, then 3 units

LenaWriter [7]

Answer:7:8

Step-by-step explanation:answer

6 0

3 years ago