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3 years ago

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A line passes through the point (2, 3) and has an undefined slope, what is the equation of the line? A) y = 2 B) y = 3 C) x = 2

Eliminate D) x = 3

1 answer:

Harman [31]3 years ago

7 0

"Undefined" slope means the line is vertical, of the form x = constant. The constant is the x-value of the point the line goes through.

C) x = 2

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El patrimonio de la empresa El Cedro es de Q187,000.00 y el pasivo es de Q69,852.00. ¿A cuanto asciende el activo?

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2 years ago

1. For the function f(x) = x2, Find the slope of the secant line joining (1,f(1)) and

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The slope of the line connecting two points (<em>a</em>, <em>b</em>) and (<em>c</em>, <em>d</em>) is

(<em>d</em> - <em>b</em>) / (<em>c</em> - <em>a</em>)

i.e. the change in the <em>y</em>-coordinate divided by the change in the <em>x</em>-coordinate. For a function <em>y</em> = <em>f(x)</em>, this slope is the slope of the secant line connecting the two points (<em>a</em>, <em>f(a)</em> ) and (<em>c</em>, <em>f(c)</em> ), and has a value of

(<em>f(c)</em> - <em>f(a)</em> ) / (<em>c</em> - <em>a</em>)

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3 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

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Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

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