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LUCKY_DIMON [66]
3 years ago
5

Dos agricultores, padre e hijo, tardan 2 horas entre los dos arar un campo. Si solo el padre tarda 6 horas. ¿Cuanto tardara el h

ijo?
Mathematics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

El hijo solo tardará 3 horas en arar el campo

Step-by-step explanation:

Aquí, se nos dice que ambos pasan dos horas para arar el campo mientras que el padre solo pasa 6 horas.

Queremos saber el tiempo que pasará el hijo si va a trabajar solo

Sea el área del campo x

La tasa de trabajo de ambos es x / 2

La tasa de trabajo del padre será x / 6 Deja que el tiempo que pasa el hijo sea y La tasa de trabajo del hijo es x / y

Sumando la tasa de trabajo del padre a la del hijo da la tasa de trabajo total de ambos Por lo tanto;

x / 6 + x / y = x / 2

Despegar x 1/6 + 1 /y = 1/2

1 / y = 1/2 - 1/6

1 / y= 1/3

entonces y = 3

El hijo solo tardará 3 horas en arar el campo.

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s
Oksi-84 [34.3K]

Answer:

a) \mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

b) From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

c) P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

P(Z\geq2.070)=1-P(Z

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

\mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

Part b

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

Part c

For this case we want this probability:

P(\bar X \geq 285)

And we can use the z score defined as:

z=\frac{\bar x -\mu}{\sigma_{\bar x}}

And using this we got:

P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

And using a calculator, excel or the normal standard table we have that:

P(Z\geq2.070)=1-P(Z

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