Lily Saves a Garden

Mathematics

1 answer:

Dvinal [7]3 years ago

4 0

Seems like you already have the answer so peace

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Put the following in order for the most area in the tails of the distribution. (a) Standard Normal Distribution (b) Student's

tensa zangetsu [6.8K]

Answer:

(b) (c) (a)

Step-by-step explanation:

Standard Normal distribution has a higher peak in the center, with more area in this región, hence it has less area in its tails.

Student's t-Distribution has a shape similar to the Standard Normal Distribution, with the difference that the shape depends on the degree of freedom. When the degree of freedom is smaller the distribution becomes flatter, so it has more area in its tails.

Student's t-Distributionwith 1515 degrees of freedom has mores area in the tails than the Student's t-Distribution with 2020 degrees of freedom and the latter has more area than Standard Normal Distribution

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2 years ago

Solve the inequality. - 16 - 4x _<0 The solution is x

Scrat [10]

Answer:

the answer for will be x≥-4

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3 years ago

You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.

postnew [5]

This problem seems to be wrong because no minimum point was found and no point of landing exists

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

Derivatives

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

The function provided in the question is

$h(x)=2x^2-12x+20$

Let's find the first derivative

$h'(x)=4x-12$

solving h'=0:

$4x-12=0$

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

To find when will the ball land, we set h=0

$2x^2-12x+20=0$