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adelina 88 [10]
3 years ago
10

Define an analog and digital signal

Computers and Technology
1 answer:
nexus9112 [7]3 years ago
7 0
<span>An analog signal is a continuous wave that changes over a time period.
</span><span>A digital signal is a discrete wave that carries information in binary form.</span>
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On average, someone with a Bachelor's degree is estimated to earn ____ times more than someone with a high school diploma.
oee [108]
The correct answer is <span>B)</span> 1.4

It can be difficult to become a high earner in the U.S without a college degree. Those with a high school diploma or less are the lowest earners on average. According to research, average yearly earnings for someone with a bachelor’s degree are $59,124 as compared to someone with a high school diploma who earns an average of $35,256 per year. Thus, at the end of the day, the amount of money does become significant.



5 0
3 years ago
Which type of website is most likely to be biased when providing information about a product?
vlabodo [156]
If the website is talking about someone or something close to the heart of the website. You wouldn't want the website you work for to talk trash about you.

3 0
3 years ago
Read 2 more answers
What is the rate constant at 25.0 ∘c based on the data collected for trial b?
Alecsey [184]
<h2><u>Answer: </u></h2>

acetone + I2 + HCl ---> iodated acetone  

Equation:

rate = k * [acetone]^x * [I2]^y * [HCl]^z  

Once we know x, y, z, we can plug in any of the trials A->D and determine k  NOTE: We can't use run E because temperature has an effect on rate. E was run at a different temperature.

The first thing to note is that do NOT have concentrations. We have volumes at a given molarity.  

Table:  

.001M I2.. ..050M HCl.. .1.0M acetone.. .water.. temp..time.. total vol  

.. ....mL.. ... ... .. .mL.. .. ... .. .. mL.. .. .. .. ..mL.. ..°C.. .sec.. .. .. .L  

A.. ...5.. .. .. .. .. ..10.. .. .. .. .. ..10.. .. .. .. ..25.. .. .25.. .130.. .. 0.05  

B.. ..10.. .. .. ... .. 10.. .. .. ... .. .10.. .. .. ... ..20.. . .25.... 249.. ..0.05  

C.. . 10.. .. .. .. .. .20.. .. .. .... .. 10.. .. ... ... .10... ..25.. ..128... .0.05  

D.. . 10.. ... .. ... ..10.. .. ... .. ... 20.. .. .. .. .. 10.. .. 25.. ..131.. ..0.05  

E.. ..10.. ... .. ... ..10.. ... ... ... ..10.. ... .. ... .20.. .. 42.8.. .38.. ..0.05  

We can translate that into molarity in solution using this formula. (molarity pure ingredient * mL used / 1000 / total volume in liters)  

 

.. .. ..I2.. .... HCl.. acetone.. temp.. ..rxn time  

.. .. ..M.. .. ...M.. .... .M.. .... ..°C.. .. .. sec  

A.. 0.0001.. 0.01..... 0.2.. .... .25... .... .130  

B.. 0.0002.. 0.01.. .. 0.2.. .. .. 25.. .. .. .249  

C.. 0.0002.. 0.02.. .. 0.2.. .. .. 25.. .. .. .128  

D.. 0.0002.. 0.01.. .. 0.4.. .. . .25.. ... ...131  

E.. 0.0002.. 0.01.. .. 0.2.. . ....42.8.. .. .. 38  

From runs B and D, we can see that rate dropped by half .

As [I2] and [HCl] were held constant and [acetone] was doubled.  

This means x=-1 in this equation  

Rate = k * [acetone]⁻¹ * [I2]^y * [HCl]^z  

Rate = k * [I2]^y * [HCl]^z  

From runs A and B  

[I2] doubles  

[HCl] remains the same  

[acetone] remains the same  

rate doubles   as [I2] doubles, rate doubles  

y = 1   rate = k * [acetone]⁻¹ * [I2]¹ * [HCl]^z  

And from runs B and C, we can see that  , As [HC] doubles, (all else equal) the rate halves.

Z = -1  

Rate = k * [I2] / ([acetone] * [HCl])  

Rearranging  

k = rate * [acetone] * [HCl] / [I2]  

From any experimental run (A-D), we can calculate k.

using A to calc k... ..k = 2600 M²/sec  

using B to calc k... . k = 2490 M²/sec  

using C to calc k... . k = 2560 M²/sec  

using D to calc k... . k = 2620 M²/sec  

NOTE.. the problem statement said to use the data from run B to calc k.

Hence Final Answer:

k = 2490 M²/sec

4 0
3 years ago
PLEASE HELP I WILL MEDAL AND FAN!
ladessa [460]
Hi!

Question 1: George W. Bush.

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Question 3: About 25 percent.

Question 4: True.

Question 5: To ensure the safety of officers and inmates and to punish those who break the rules.

Question 6: I believe it is false.

Question 7: True.

Question 12: John F. Kennedy.

Question 13: I believe it's False.

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That's all I know!
4 0
4 years ago
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Computer A uses Stop and Wait ARQ to send packets to computer B. If the distance between A and B is 40000 km, the packet size is
e-lub [12.9K]

The time that it takes the computer to receive acknowledgment for a packet is 0.1667 seconds. The time it takes to send out a packet is  4 x 10⁻³seconds

<h3>1 The acknowledgment time for the packet</h3>

speed =  2.4x108m/s

Distance = 40000 km,

Time = distance/ speed

= 40000 x10³/ 2.4x10⁸m/s

= 0.1667

The time that it take is 0.1667 seconds.

b. Number of bytes = 5000

5000x 8 = 40000bits

10 mbps = 10000 kbps

10000 kbps = 10000000

packet size / bit rate = 40000/10000000

= 4 x 10⁻³seconds to send a packet out

Read more on computer bandwith here: brainly.com/question/27020560

4 0
2 years ago
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