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3 years ago

What is the minimum value of the function g(x) = x2 - 6x – 12? -21 3-21 3 3+21

Mathematics

1 answer:

balandron [24]3 years ago

4 0

Answer:

- 21

Step-by-step explanation:

The minimum value occurs at the vertex of the function

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

To obtain this form use the method of completing the square

Given

f(x) = x² - 6x - 12

add/ subtract ( half the coefficient of the x- term)² to x² - 6x

f(x) = x² + 2(- 3)x + 9 - 9 - 12

= (x - 3)² - 21

with vertex = (3, - 21 )

The minimum is the value of k, that is minimum value = - 21

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Step-by-step explanation:

Hello!

When all other things remain the same, which of the following statements about the width of the interval is correct?

A. The interval will be narrower if 15 men are used in the sample.

B. The interval will be wider if 15 men are used in the sample.

C. The interval will be narrower if 5 men are used in the sample.

D. The interval will be narrower if the level is increased to 99% confidence.

E. The interval will be wider if the level is decreased to 90% confidence.

Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.

The general structure for the CI is "point estimate"±" margin fo error"

Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)

The margin of error of the interval is:

d= $t_{n-1;1-\alpha /2}$ * (Sd/n)

1) The sample size changes, all other terms of the interval stay the same.

As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.

↓d= $t_{n-1;1-\alpha /2}$ * (Sd/↑n)

↑d= $t_{n-1;1-\alpha /2}$ * (Sd/↓n)

Correct option: A. The interval will be narrower if 15 men are used in the sample.

2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:

↓d= ↓ $t_{n-1;1-\alpha /2}$ * (Sd/n)

↑d= ↑ $t_{n-1;1-\alpha /2}$ * (Sd/n)

I hope it helps!