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2 years ago

Which inequality represents the sentence below?

Mathematics

2 answers:

nikitadnepr [17]2 years ago

8 0

The difference of a number P and one and four tenths is more than nine and seventy-eight hundredths.

P - 1.4 > 9.78

Alona [7]2 years ago

3 0

Answer:

A. $P-1.4>9.78$

Step-by-step explanation:

We have been given a sentence. We are asked to find the inequality that represents the given sentence.

Sentence:

The difference of a number and one and four tenths is more than nine and seventy-eight hundredths.

Let the number be P.

The difference of number and one and four tenths would be $P-1.4$ .

The difference both numbers is more than nine and seventy-eight hundredths. We can represent this information in an inequality as:

$P-1.4>9.78$

Therefore, option A is the correct choice.

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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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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:

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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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v-10=-3 I don't get how this would get a negative answer I don't get it and my teacher sucks at explaining these so plz help!!!!

o-na [289]

V-10=-3

add 10 by both sides (opposite of -10)

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