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4 years ago

1. Practicando: expresa en lenguaje algebraico.

Mathematics

1 answer:

Natasha_Volkova [10]4 years ago

8 0

Answer:

Part 1) $3n$

Part 2) $4n$

Part 3) $\frac{2}{3}n$

Step-by-step explanation:

The question in English is

Expressed in algebraic language.

Symbolically write the following expressions:

Three times a number:

Four times a number:

Two-thirds of a number

Let

n ----> the number

Part 1) we have

Three times a number:

we know that

The algebraic expression is equal to multiply 3 by the number n

$3n$

Part 2) we have

Four times a number:

we know that

The algebraic expression is equal to multiply 4 by the number n

$4n$

Part 3) we have

Two-thirds of a number

we know that

The algebraic expression is equal to multiply the fraction 2/3 by the number n

$\frac{2}{3}n$

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Gre4nikov [31]

Answer:

a) There is a 99.99997% probability that at least one valve opens.

b) There is a 22.62% probability that at least one valve fails to open.

Step-by-step explanation:

For each valve, there are only two possible outcomes. Either it open, or it does not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.95, n = 5$

(a) What is the probability that at least one valve opens?

Either no valves open, or at least one opens. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

$P(X \geq 1) = 1 - P(X = 0)$

$P(X = 0) = C_{5,0}.(0.95)^{0}.(0.05)^{5} = 0.0000003$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000003 = 0.9999997$

There is a 99.99997% probability that at least one valve opens.

(b) What is the probability that at least one valve fails to open?

Either all valves open, or at least one does not open. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \leq 4)$

$P(X \leq 4) = 1 - P(X = 5)$

$P(X = 5) = C_{5,5}.(0.95)^{5}.(0.05)^{0} = 0.7738$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.7738 = 0.2262$

There is a 22.62% probability that at least one valve fails to open.

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erastovalidia [21]

Answer: $y=\frac{-x}{w-3}$

Step-by-step explanation:

To solve for y, we want to isolate y.

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Now, we know that $y=\frac{-x}{w-3}$ .

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Answer:

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