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Marizza181 [45]
3 years ago
7

Al obtener el valor de lo que costará la fiesta de graduación para tu grupo te das cuenta que el lugar que decidieron alquilar p

ara la fiesta es la Hacienda La Providencia, que cuenta con diferentes salones para el evento: uno tiene capacidad para 1,500 personas a 800 personas. El precio del boleto por persona varía según el número de asistentes que decidan comprar el boleto. La siguiente es la función que moldea el comportamiento: f(x)=〖-2x〗^2+80x+300
Mathematics
1 answer:
dybincka [34]3 years ago
8 0

Explicación paso a paso:

Debemos buscar el costo de cada boleto y la cantidad de personas que deben ir.

Dada la función que da forma al comportamiento expresado como;

f (x) = (-2x) ^ 2 + 80x + 300 donde;

x es el costo de cada boleto

f (x) es el número de personas que deben ir

Para obtener x, usaremos la expresión;

x = -b / 2a

De la ecuación cuadrática dada, a = -2, b = 80 yc = 300

Sustituir;

x = -80/2 (-2)

x = -80 / -4

x = 80/4

x = 20

Por lo tanto, el boleto cuesta $ 20.

Lo siguiente es obtener la cantidad de personas que deben ir. Para conseguirlo, sustituiremos x = 20 en la expresión modelada como se muestra;

f (x) = (-2x) ^ 2 + 80x + 300

f (20) = (-2 (20)) ^ 2 + 80 (20) +300

f (20) = (-40) ^ 2 + 1600 + 300

f (20) = 1600 + 1600 + 300

f (20) = 3500

Por lo tanto, 3500 personas tienen que ir

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Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

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Its sample mean is X[bar]₁= 69 feet

And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.

The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

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α: 0.05

The statistic to use to compare these two populations is a pooled Z test

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } }

Z ≈ N(0;1)

Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397

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If Z_{H_0} > -1.648 , then you do not reject the null hypothesis.

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Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

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