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3 years ago

The function y = 3 squared - x - 3 is graphed only over the domain of {x | –8 < x < 8}. What is the range of the graph?

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer: Y < = 144

Step-by-step explanation:

Given that the function y = 3 squared - x - 3 is graphed only over the domain of {x | –8 < x < 8}

That is, y = 3x^2 - 3

Since x is greater than -8, the minimum value of x will be -7.

Also, x is less than 8, the maximum value of x will be 7

Substitute the two values into the given function.

When x = -7

Y = 3(-7)^2 - 3

Y = 3(49) - 3

Y = 147 - 3

Y = 144

When x = 7

Y = 3(7)^2 - 3

Y = 3(49) - 3

Y = 147 - 3

Y = 144

The range of the graph is therefore, Y less than or equal to 144. That is,

Y < = 144

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Please help! I don't understand how to solve this problem

Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

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The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

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n is the sample size.

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Assuming an uniform distribution, the standard deviation is given by:

$S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547$

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

$M = z\frac{\sigma}{\sqrt{n}}$

$0.006 = 1.96\frac{1.1547}{\sqrt{n}}$

$0.006\sqrt{n} = 1.96 \times 1.1547$

$\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}$

$(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2$

n = 142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

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Encuentra la ecuacion de la parabola con vertice en el origen y foco en las cordenadas 0,1en su forma ordinaria y general

marusya05 [52]

Answer:

$x^{2}=4y$

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Step-by-step explanation:

Forma ordinaria

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$(x-h)^{2}=4p(y-k)$ (1)

Donde:

(h,k) es la coordenda del vértice, en nuestro caso (0,0) ya que está en el origen.
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Por lo tanto h = 0, k = 0 y p = 1.

Remplazando estos valores en la ecuación de la parábola, tenemos:

$(x-0)^{2}=4*1(y-0)$

$x^{2}=4y$ (2)

Forma general

La forma general de una parábola esta dada por la siquiente ecuación

$y=ax^{2}+bx+c$

Reordenando la ecuación (2) tenemos:

$y=\frac{1}{4}x^{2}$

Espero esto te haya ayudado!

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