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4 years ago

Two quadratic functions are shown.

Mathematics

1 answer:

algol [13]4 years ago

7 0

Answer: the function that has the smaller minimum is g(x), and the cordinates are (0,3)

Step-by-step explanation:

We have a function for f(x) and a table for g(x)

first, quadratic functions are symmetrical.

This means that if the minimum/maximum is located at x = x0, we will have that:

f(x0 + A) = f(x0 - A)

For any real value of A.

Then when we look at the table, we can see that:

g(-1) = 7

g(0) = 3

g(1) = 7

then the minimum of g(x) must be at x = 0, and we can see that the minimum value of g(x) is 3.

Now let's analyze f(x).

When we have a quadratic equation of the shape.

y = a*x^2 + b*x + c

the minimum/maximum will be located at:

x = -b/2a

In our function we have:

a = 3

b = 6

then the minimum is at:

X = -6/2*3 = -1

f(-1) = 3*(-1)^2 + 6*-1 + 7 = 3 - 6 + 7 = 3 + 1 = 4

Then the function that has the smaller minimum is g(x), and the cordinates are (0,3)

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"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

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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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

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