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3 years ago

The major difference between a template and another document is in.

1 answer:

4 0

Templates in pandadoc are used for generic content that you intend on using multiple times, while documents are used for specific information. In order to send a document, you must first creat it from an existing template.

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PLEASE I NEED HELP FAST

tankabanditka [31]

Answer:

your answer will be src=

7 0

3 years ago

We want to implement a data link control protocol on a channel that has limited bandwidth and high error rate. On the other hand

Korolek [52]

Answer:

Selective Repeat protocols

Explanation:

It is better to make use of the selective repeat protocol here. From what we have here, there is a high error rate on this channel.

If we had implemented Go back N protocol, the whole N packets would be retransmitted. Much bandwidth would be needed here.

But we are told that bandwidth is limited. So if packet get lost when we implement selective protocol, we would only need less bandwidth since we would retransmit only this packet.

6 0

3 years ago

What is the name of the item that supplies the exact or near exact voltage at the required wattage to all of the circuitry insid

Karo-lina-s [1.5K]

Answer:

<em>A voltage regulator.</em>

Explanation:

<em>A voltage regulator, controls the output of an alternating current or a direct current (depending on the design), allowing the exact amount of voltage or wattage to be supplied to the computer hardware</em>. This device sometimes uses a simple feed-forward design or may include negative feedback. The two major types of voltage regulator are based on either the electromechanical or electronic components.

<em>The electronic types were based on the arrangement of resistor in series with a diode or series of diodes, and the electromechanical types are based on coiling the sensing wire to make an electromagnet. </em>

4 0

4 years ago

Read the following code used to calculate the weight of an item with two-pound packaging:

Veronika [31]

When there is an error in the code, the function that should be used is A. float(); weight requires decimals

Computer code, or a set of instructions or a system of rules defined in a specific programming language, is a term used in computer programming (i.e., the source code). It is also the name given to the source code after a compiler has prepared it for computer execution.

A float is a number with a decimal place since it is a floating-point number. When greater precision is required, floats are employed. The provided value is transformed into a floating point number by the float() function. In this case, the code is used to calculate the weight of an item and the float function will be vital.

Learn more about programs and code on:

brainly.com/question/22654163

#SPJ1

7 0

1 year ago

Note: You can use a word document to write your answers and copy-paste your answer to the area specified. a. (5 points) Convert

MrMuchimi

Answer:

$EA9_{16} = 3753$

$CB2_{16} = 3250$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(1010 1000 1011 1000 1110 1101)_2 = 11057389$

$(1011 1110 1101 1011 1010)_2 = BEDBA$

$(1010 1000 1011 1000 1110 1101)_2 = A8B8ED$

$74510_8= 221416$

$67210_8 = 203212$

Explanation:

Solving (a): To base 10

$(i)\ EA9_{16$

We simply multiply each digit by a base of 16 to the power of their position.

i.e.

$EA9_{16} = E * 16^2 + A * 16^1 + 9 * 16^0$

$EA9_{16} = E * 256 + A * 16 + 9 * 1$

In hexadecimal

$A = 10; E = 14$

So:

$EA9_{16} = 14 * 256 + 10 * 16 + 9 * 1$

$EA9_{16} = 3753$

$(ii)\ CB2_{16}$

This gives:

$CB2_{16} = C * 16^2 + B * 16^1 + 2 * 16^0$

$CB2_{16} = C * 256 + B * 16 + 2 * 1$

In hexadecimal

$C = 12; B =11$

So:

$CB2_{16} = 12 * 256 + 11 * 16 + 2 * 1$

$CB2_{16} = 3250$

Solving (b): To base 10

$(i)\ (1011 1110 1101 1011 1010)_2$

We simply multiply each digit by a base of 2 to the power of their position.

i.e.

$(1011 1110 1101 1011 1010)_2 = 1 * 2^{19} + 0 * 2^{18} + 1 * 2^{17} + 1 * 2^{16} +1 * 2^{15} + 1 * 2^{14} + 1 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 1 * 2^{10} + 0 * 2^9 + 1 * 2^8 +1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0$

$(1011 1110 1101 1011 1010)_2 = 781754$

$(ii)\ (1010 1000 1011 1000 1110 1101)_2$

$(1010 1000 1011 1000 1110 1101)_2 = 1 * 2^{23} + 0 * 2^{22} + 1 * 2^{21} + 0 * 2^{20} +1 * 2^{19} + 0 * 2^{18} + 0 * 2^{17} + 0 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 1 * 2^{13} + 1 * 2^{12} +1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 0 * 2^8 + 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 1*2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0$