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1 year ago

Ayudeme a resolver y poner signo positivo y negativo

Mathematics

1 answer:

Lina20 [59]1 year ago

4 0

Debido a restricciones de extensión y la características del ejercicio, recomendamos leer la explicación de esta pregunta para mayores detalles sobre la adición de números enteros.

<h3>¿Cuáles son los resultados de cada suma?</h3>

En este ejercicio tenemos un grupo de sumas con números enteros positivos y negativos, en las cuales se prueba la capacidad del estudiante para realizar varias operaciones en serie (adición, sustracción) y comprender las diferencias entre números positivos, negativos y neutros. Ahora procedemos a determinar el resultado de cada una de las expresiones:

20 + 50 + 30 + 7 = 107

30 + 5 + 2 = 37

- 200 - 50 - 70 - 8 = - 328

- 500 + 100 - 20 + 50 = - 370

10 - 5 = 5

20 + 50 - 25 - 10 = 35

- 100 + 20 = - 80

- 30 + 5 + 4 - 20 + 8 = - 33

- 258 + 8 = - 250

- 10 + 20 + 520 - 100 + 8 = 438

- 20 - 5 - 42 + 3 = - 64

1000 - 200 + 50 + 30 - 45 + 75 - 87 + 90 + 50 - 100 + 50 - 10 = 903

- 400 + 500 - 200 - 50 + 48 + 8 - 47 - 50 = - 191

300 + 20 - 50 + 30 - 84 + 35 - 7 + 20 - 40 + 10 - 45 + 65 + 8 - 55 = 207

800 + 50 - 69 + 8 - 35 + 85 - 54 + 40 + 85 + 74 - 32 - 8 + 65 - 27 = 982

Para aprender más sobre sumas: brainly.com/question/1456841

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2222222222222222222222 x 0

tangare [24]

Answer: 0

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Critical Thinking: Empirical/Quantitative Skills

aliya0001 [1]

Answer:

1. 0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. 0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. 0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

Step-by-step explanation:

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume that there is a 0.905 probability that a passenger with a ticket will show up for the flight.

This means that $p = 0.905$

Also assume that the airline sells 200 tickets

This means that $n = 200$

Question 1:

Exactly, so we can use the P(X = x) formula, to find P(X = 180).

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

$P(X = 180) = C_{200,180}.(0.905)^{180}.(0.095)^{20} = 0.0910$

0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. When 200 tickets are sold, calculate the probability that at most 180 passengers show up for the flight.

Now we have to use the approximation.

Mean and standard deviation:

$\mu = E(X) = np = 200*0.905 = 181$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.905*0.095} = 4.15$

Using continuity correction, this is $P(X \leq 180 + 0.5) = P(X \leq 180.5)$ , which is the p-value of Z when X = 180.5. Thus

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{180.5 - 181}{4.15}$

$Z = -0.12$

$Z = -0.12$ has a p-value of 0.4522.

0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. When 200 tickets are sold, calculate the probability that more than 180 passengers show up for the flight.

Complementary event with at most 180 passengers showing up, which means that the sum of these probabilities is 1. So

$p + 0.4522 = 1$

$p = 1 - 0.4522 = 0.5478$

0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

4 0

3 years ago

Two numbers whose sum is 29 and their difference is 15

n200080 [17]

Answer:

22 and 7

Step-by-step explanation:

Let x and y be the two numbers we are looking for:

The statement “their sum is 29” means x + y = 29

The statement “their difference is 15” means x - y = 15

we get this system:

x + y = 29 (1)

x - y = 15 (2)

Let’s solve it :

(1) + (2) ⇒ x + y + x - y = 29 + 15 ⇒ 2x = 44 ⇒ x = 44/2 = 22

Now we replace x by 22 in equation (1) :

22 + y = 29 ⇒ y = 29 - 22 = 7

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2 years ago

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Vladimir [108]

B is the answer how you do good

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3 years ago

This really hard help please

bonufazy [111]

Answer:

is it 4.8, I don't really know

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3 years ago