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2 years ago

) What does sampling mean?

Computers and Technology

1 answer:

lubasha [3.4K]2 years ago

6 0

In terms of music taking one song and incorporating it into your own song

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It allows you to create tiled copies of an object while tweaking the variables

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Yet another variation: A better packet switched network employs the concept of acknowledgment. When the end user’s device receiv

dlinn [17]

Answer:

a. see explaination

b. 0.632

Explanation:

Packet switching is a method of grouping data that is transmitted over a digital network into packets. Packets are made of a header and a payload.

See attachment for the step by step solution of the given problem.

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3 years ago

Use the following cell phone airport data speeds (Mbps) from a particular network. Find P10. 0.1 0.1 0.3 0.3 0.3 0.4 0.4 0.4 0.

sasho [114]

Answer:

$P_{10} =0.1$

Explanation:

Given

$0.1, 0.1, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.6, 0.7, 0.7, 0.7, 0.8, 0.8$

Required

Determine $P_{10}$

$P_{10}$ implies 10th percentile and this is calculated as thus

$P_{10} = \frac{10(n+1)}{100}$

Where n is the number of data; n = 14

$P_{10} = \frac{10(n+1)}{100}$

Substitute 14 for n

$P_{10} = \frac{10(14+1)}{100}$

$P_{10} = \frac{10(15)}{100}$

Open the bracket

$P_{10} = \frac{10 * 15}{100}$

$P_{10} = \frac{150}{100}$

$P_{10} = 1.5th\ item$

This means that the 1.5th item is $P_{10}$

And this falls between the 1st and 2nd item and is calculated as thus;

$P_{10} = 1.5th\ item$

Express 1.5 as 1 + 0.5

$P_{10} = (1 +0.5)\ th\ item$

$P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item$

From the given data; $1st\ item = 0.1$ and $2nd\ item = 0.1$

$P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item$ becomes

$P_{10} =0.1 +0.5(0.1 - 0.1)$

$P_{10} =0.1 +0.5(0)$

$P_{10} =0.1 +0$

$P_{10} =0.1$

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3 years ago

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Licemer1 [7]

Answer:

c. all of these choices

Explanation:

all of them are correct. i took a quiz on this like 2 days ago and that was the answer.

<em>hope this helpsssssssss!!!!!!!!!!! :):):)</em>

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3 years ago

True or False: It is illegal to park in a location that you block or create a hazard for other vehicles.

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The answer to question is true

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