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kotykmax [81]
3 years ago
9

Explain the term software dependability. Give at least two real-world examples which further elaborates

Computers and Technology
1 answer:
Elena L [17]3 years ago
3 0

Answer:

Explanation:In software engineering, dependability is the ability to provide services that can defensibly be trusted within a time-period. This may also encompass mechanisms designed to increase and maintain the dependability of a system or software.Computer software is typically classified into two major types of programs: system software and application software.

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The incompatibilities in speed between the various devices and the CPU make I/O synchronization difficult, especially if there a
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In a buffer

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Items stored at the buffer helps to reduce the The incompatibilities in speed between the various devices and the CPU.

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How I can learn writing code?
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The following are reasons why many organizations are trying to reduce and right-size their information foot-print by using data
blagie [28]

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C.<em> improve data quality and reduce redundancies, reduce increased and staggering storage management costs</em>

<em></em>

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3 years ago
How do you go about inserting additional worksheets into a workbook?
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The correct answer is D. All of the methods listed above are ways of inserting new sheets into a workbook.
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3 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
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