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dybincka [34]
2 years ago
12

Escribe una ecuación de la recta que pasa por el punto (5, –8) con pendiente 5.

Mathematics
1 answer:
Nana76 [90]2 years ago
4 0

Considerando la expresión de una ecuación lineal, la ecuación de la recta que pasa por el punto (5, –8) con pendiente 5 es y= 5x - 33.

<h3>Ecuación lineal</h3>

Una ecuación lineal o línea se puede expresar en la forma y = mx + b

donde

  • x y son coordenadas de un punto.
  • m es la pendiente.
  • b es la ordenada al origen y representa la coordenada del punto donde la línea cruza el eje y.

La ecuación lineal se puede expresar también mediante la ecuación punto-pendiente de la recta, que se plantea si se conoce la pendiente de la recta y cualquiera de sus puntos. Con ello queda determinada la recta, conociendo la pendiente  “m” y un punto  (x1, y1) dado:

y - y1= m(x -x1)

<h3>Ecuación de la recta en este caso</h3>

En este caso, la recta que pasa por el punto (5, –8) con pendiente 5.

Reemplazando en la ecuación punto-pendiente de la recta:

y - (-8)= 5(x -5)

Resolviendo:

y + 8= 5(x -5)

Aplicando propiedad distributiva:

y + 8= 5x - 5×5

y + 8= 5x - 25

Aislando la variable "y":

y= 5x - 25 - 8

<u><em>y= 5x - 33</em></u>

Finalmente, la ecuación de la recta que pasa por el punto (5, –8) con pendiente 5 es y= 5x - 33.

Aprende más sobre ecuación de una recta:

brainly.com/question/25243069

brainly.com/question/19260315

brainly.com/question/24766917

#SPJ1

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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

Z = \frac{X - \mu}{s}

X = 205

Z = \frac{X - \mu}{s}

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

Z = \frac{X - \mu}{s}

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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3 years ago
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Y_Kistochka [10]
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Vladimir [108]

Answer:

The number of different 7 card hands possibility is  133784560

Step-by-step explanation:

The computation of the number of different 7 card hands possibility is shown below:

Here we use the combination as the orders of choosing it is not significant

So,

The number of the different hands possible would be

= ^{52}C_7

= 52! ÷ (7! × (52 - 7)!)

= 133784560

Hence, the number of different 7 card hands possibility is  133784560

4 0
3 years ago
Is 1000 a term in the sequence 50,65,80,95
Alex Ar [27]
Yes it is

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EXPLANATION:

the nth term would be 15n+35
because the terms go up by 15 and you add 35 to get to the term.


1. 2. 3. 4.
50. 65. 80. 95

1000- 35= 965


965/15=41

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2 years ago
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