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Free_Kalibri [48]
3 years ago
13

The following situation calls for a significance test. State the appropriate null hypothesis

_0" id="TexFormula1" title="H_0" alt="H_0" align="absmiddle" class="latex-formula"> and alternative hypothesis H_a.
Be sure to define your parameter.
When ski jumpers take off, the distance they fly varies considerably depending on their speed, skill, and wind conditions. Event organizers must position the landing area to allow for differences in the distances that the athletes fly. For a particular competition, the organizers estimate that the variation in distance flown by the athletes will be σ = 10 meters. An experienced jumper thinks that the organizers are underestimating the variation.
Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

H_{0}: \sigma = 10\\

H_{a}: \sigma > 10\\        

Step-by-step explanation:

We are given the following in the question:

The organizers estimate that the variation in distance flown by the athletes will be σ = 10 meters.

This will be stated by the null hypothesis.

H_{0}: \sigma = 10\\

It was claimed that the organizers are underestimating the variation and the variation in distance is greater than 10.

This will be stated by the alternate hypothesis.

H_{a}: \sigma > 10\\

The parameter for the given hypothesis is standard deviation, \sigma.

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Karma is building a rectangular table with an area of 18,000 cm?. He wants to put wood trim around the four edges. What is the s
jok3333 [9.3K]

The shortest length is given by the function for the perimeter of the

rectangular table.

  • The shortest length of trim he could use is <u>536.66 cm</u>
<h3>Method used for finding the shortest length of trim</h3>

The given parameter;

Area of the rectangular table Karma is building, <em>A</em> = <u>18,000 cm²</u>

Required:

The shortest length of trim he could use which he wants to put around the four edges.

Solution:

Let <em>l</em> represent the length of the table, and let <em>w</em> represent the width, therefore;

Perimeter of the table, <em>P</em> = 2·l + 2·w

Area, <em>A</em> = l × w

Which gives;

18,000 = l × w

l = \dfrac{18,000}{w}

Which gives;

P = 2 \cdot \dfrac{18,000}{w} + 2 \cdot w

At the minimum point, we have;

\dfrac{d}{dw} P = \dfrac{d}{dw} \left(2 \cdot \dfrac{18,000}{w} + 2 \cdot w\right)= \mathbf{\dfrac{2 \cdot w^2 - 36,000}{w^2} }= 0

Which gives;

2·w² - 36,000 = w² × 0 = 0

2·w² = 36,000

w^2 = \dfrac{36,000}{2} = 18,000

The width of the rectangular table, <em>w</em> = √(18,000)

Length \ of \ the \ table\ l = \dfrac{18,000}{\sqrt{18,000} } = \sqrt{18,000}

Therefore;

The perimeter of the table, P ≈ 2 × √(18,000) + 2 × √(18,000) ≈ 536.656

The length of trim required = The perimeter of the rectangular table, <em>P</em>

Therefore;

  • The shortest length of the trim he could use, given to the nearest hundredth is <u>536.66 cm</u>

Learn more about area and perimeter of a figure here:

brainly.com/question/9135929

4 0
3 years ago
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