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Tatiana [17]
3 years ago
6

un solar tiene forma de 3/4 de círculo unido con un triángulo rectángulo , En este solar va a ser utilizado para sembrar clavele

s. el agricultor tiene que comprar el abono y adecuar el techo para mejor conservación de los claveles. el sabe que el metro cuadrado de abono cuesta $8016 Y que cada metro cuadro de techo cuesta $2974. Si el solar tiene un radio de 5.9 metros, entonces el valor total a intervenir para la adecuación de este es?​

Mathematics
1 answer:
Stolb23 [73]3 years ago
7 0

Answer:

$1092671.191

Step-by-step explanation:

<u>Cálculo de Areas</u>

El área de un círculo de radio r se obtiene con la fórmula:

A_c=\pi r^2

El área de un triángulo de base b y altura h, perpendicular a la base es:

A_t=\frac{bh}{2}

Si el triángulo es rectángulo, b y h son los catetos

El solar tiene una forma tal como se ilustra en la figura anexa. Para averiguar el costo del abono y del techo, debe calcularse primero el área total a cultivar. Primero se determina el área de 3/4 de círculo de radio 5.9 metros

A_c=\frac{3}{4}\pi r^2

A_c=\frac{3}{4}\pi (5.9)^2

A_c=82.0191302\ m^2

Por su parte, el triángulo mostrado, tiene ambos catetos iguales al radio de círculo, por lo que su área será

A_t=\frac{r^2}{2}

A_t=\frac{(5.9)^2}{2}

A_t=17.405\ m^2

El área total del solar es la suma de ambas:

A_s=82.0191302\ m^2+17.405\ m^2

A_s=99.4241302\ m^2

El metro cuadrado de abono cuesta $8016. El costo de abono es

C_a=99.4241302\ m^2*8016=\$796983.8277

El metro cuadrado de techo cuesta $2974. El costo del techo es

C_t=99.4241302\ m^2*2974=\$295687.3632

El costo total es la suma de los dos anteriores:

Total=\$796983.8277+\$295687.3632

Total=\$1092671.191

El valor total a intervenir para la adecuación del solar es $1092671.191

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Answer:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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".

Solution to the problem

Part a

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

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

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

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