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Cloud [144]
2 years ago
11

Which of these is not a valid formlayout in Microsoft Access?​

Computers and Technology
1 answer:
Ksivusya [100]2 years ago
3 0
There’s nothing provided for me to chose from
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need the answer ASAP. I’ll mark brainliest <br><br> they’re all the same options for each sentence.
KengaRu [80]

Answer:

Explanation:

pop(): Remove an item from the end of an array  

push(): Add items to the end of an array  

shift(): Remove an item from the beginning of an array  

unshift(): Add items to the beginning of an array

4 0
2 years ago
The processing of data in a computer involves the interplay between its various hardware components.
Ronch [10]

True.

Data processing involves the conversion of raw data and the flow of data through the Central Processing Unit and Memory to output devices. Each CPU in a computer contains two primary elements: the Arithmetic Logic Unit (ALU) and the control unit. The Arithmetic Logic Unit performs complex mathematical calculations and logical comparisons. On the other hand, the control unit accesses computer instructions, decodes them, and controls the flow of data in and out of the Memory, ALU, primary and secondary storage, and various other output devices.  


8 0
3 years ago
What does "bbl" mean? my friend uses text slang all the time and i can never understand it unless it's lol.? please help!
Mariana [72]
<span>It means be back later :)</span>
5 0
3 years ago
3. The wildlife department has been feeding a special food to rainbow trout finger lings in a pond. A sample of the weight of 40
andrey2020 [161]

Answer:

1)  402.7 grams. This estimate is called the sample mean.

2)  (399.11, 406.29)

3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.

I am 99% sure that the value lies between 399.11 grams and 406.29 grams.

Explanation:

sample size (n) = 40, the mean weight (x)= 402.7 grams and the standard deviation (σ)=8.8 grams

1) The point estimated mean weight of the population is 402.7 grams. This estimate is called the sample mean.

2) c = 99% = 0.99

α = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2} =0.005.

The z score of 0.005 corresponds with the z score of 0.495 (0.5 - 0.005).

z_\frac{\alpha }{2} =2.58.

The margin of error (e) = z_\frac{\alpha }{2}*\frac{\sigma}{\sqrt{n} }  =2.58*\frac{8.8}{\sqrt{40} } =3.59

The confidence interval = x ± e = 402.7 ± 3.59 = (399.11, 406.29)

3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.

I am 99% sure that the value lies between 399.11 grams and 406.29 grams.

8 0
3 years ago
What is a benefit of using client/server network?
ANTONII [103]

The answer is centralization

A client-server network is a server which acts as one powerful computer that provides information to other sub-ordinate workstations called the client. The biggest benefit of this setup is centralization or the central management of the server. Only a single server functions as a host that all clients request and use. Having a single centrally managed server is cost effective and is easily manageable. This centralized control can also be used to deploy processes, allocate user resources and delegate access permissions easily and effectively.

3 0
2 years ago
Read 2 more answers
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