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3 years ago

What must be done before using ArrayLists? (This is Java Programming btw)

Computers and Technology

1 answer:

Viktor [21]3 years ago

5 0

Answer:

to create an ArrayList , you declare an Arraylist variable and call the ArrayList constructor to instantiate and ArrayList object and assign it to the variable

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olga_2 [115]

It seems like a great county to visit, with many great views and tourist spots throughout.

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3 years ago

USB 3.1 offers an improvement over its predecessors by enabling a maximum data transfer rate of up to:

ch4aika [34]

Answer:

10 Gbps

Explanation:

The full form USB is Universal Serial Bus. It is used to connect to electronics as well as computers. The 3.1 USB is a recent and advance version of the USB standard. It was released in the year 2013. It can transfer data speed up to a speed of 10 Gbps.

It USB 3.1 can use the USB-C connector types and other connector types as well.

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3 years ago

Based on your understanding of Moore’s Law, what do you predict will be the next trend in the evolution of computers?

miss Akunina [59]

Actually, the answer is (B) Computers will continue to become smaller in size and become faster. For example, google i9-9900k mini itx pc build. You can have an insanely powerful pc and it be very small.

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3 years ago

What is the new file system available in windows server 2012?

exis [7]

Windows 2012 introduced a proprietary File system to success NTFS with ReSF (Resilient File System). There are many improvements made over NTFS as larger disks sizes, better quota management, additional attributes etc. also, the speed has significantly increased. There are many more improvement that you can simply Google.

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4 years ago

After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease

umka2103 [35]

Answer:

The probability that you have the disease, given that your test is positive is ≈ 0.0098

Explanation:

This is a conditional probability problem.

Let P(A|B) denote the conditional probability of A given B and it satisfies the equation

(1) P(A|B) = P(A) × P(B|A) / P(B)

We have the the probabilities:

P(Testing Positive | Having Disease) =0.99

P(Testing Negative | Not Having Disease) =0.99

P(Testing Positive | Not Having Disease) = 1-0.99=0.01

P(Having Disease) = 0.0001 (striking only one in 10,000 people)

P(Not Having Disease)= 1 - 0.0001 = 0.9999

We can calculate:

P(Testing Positive) =

P(Having Disease) × P(Testing Positive | Having Disease) + P(Not Having Disease) × P(Testing positive | Not Having Disease ) = 0.0001×0.99 + 0.9999×0.01 =0.010098

from (1) we have the equation:

P(Having Disease|Testing Positive)=P(Having Disease) × P(Testing Positive | Having Disease)/ P(Testing Positive) = 0.0001×0.99/0.010098≈0.0098

Thus, the probability that you have the disease, given that your test is positive is ≈ 0.0098

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4 years ago