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4 years ago

What percent of team B’s mean absolute deviation is the difference in the means?

Mathematics

2 answers:

anastassius [24]4 years ago

7 0

Answer:

Step-by-step explanation:

poizon [28]4 years ago

5 0

<h2>The difference between the mean time is about 16 times the mean absolute deviation of the data sets.</h2>

Step-by-step explanation:

We know the mean and absolute mean deviation of two team in a tables as:

Team A Team B

Mean 127.9s 127.4s

Mean absolute deviation 0.26s 0.23s

The difference of mean between two team A and B = 0.5 s

Also,

The difference of mean absolute deviation = 0.03 s

Now, let the difference of the mean times be equal to n times the difference of the mean absolute deviation of two teams.

0.5 = n x 0.03

n = 16.667 (Approx 16)

Therefore, the difference between the mean time is about the 16 times the absolute deviation of the data sets.

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3 years ago

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3 years ago

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<h3>Given information-</h3>

The triangle for the given problem is shown in the image below.

Form the figure the length of the each side is $16 \sqrt{3}$ units.

As all the sides are equal thus the $\Delta MNO$ is a equilateral triangle in which the height of the divides the triangle into two equal part of the length $8\sqrt{3}$ at point <em>R.</em>

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Now in the $\Delta MRN$ , the length of the hypotenuse is $16 \sqrt{3}$ units and the length of the base is $8\sqrt{3}$ units. Let <em>h </em>is the height of the triangle thus by the Pythagoras theorem,

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Solve for <em>h,</em>

<em />

<em /> $\begin{aligned}h^2 &=(16\sqrt{3})^2 -(8\sqrt{3})^2\\h^2 &=16\times16\times3 -8\times8\times3\\h^2 &=576\\h &=\sqrt{576}\\h &=24\\\end$ <em />

<em />

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Learn more about the equilateral triangle here;

brainly.com/question/4268382

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