Which graph represents the function y = 2/3x-2

Mathematics

1 answer:

Alina [70]3 years ago

7 0

Answer:

Step-by-step explanation:

so the y-int is where the point is on the y-axis (0,X). Or, in this situation-2.

for slope, you just need to do rise over run, or up and down first then, left and right. If I do that for C, I go 2 up and 3 right. So the slope os 2/3 (P.S. up and right is positive, and down and left is negative)

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Snezhnost [94]

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3 years ago

A reflection across the x-axis keeps the x-coordinate the same but flips the y-coordinate. What do you think a reflection across

Studentka2010 [4]

Answer: A reflection across the x-axis keeps the x-coordinates the same but flips the signs of the y-coordinates. So, it should be the opposite for a reflection across the y-axis. The y-coordinates remain the same, but the signs of the x-coordinates change.

Step-by-step explanation

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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)

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