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2 years ago

15

Which of the following can you NOT apply for at any FLHSMV office? A. Certificate of title B. License plates C. Vehicle registra

tion D. Vehicle retrieval

Computers and Technology

2 answers:

tia_tia [17]2 years ago

8 0

Answer:A

Explanation:

MArishka [77]2 years ago

7 0

The answer is D

Vehicle retrieval

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Use the single-server drive-up bank teller operation referred to in Problems 1 and 2 to determine the following operating charac

elena-s [515]

Answer:

This question is incomplete, here's the complete question:

1. Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4 customers per minute. 3. Use the single-server drive-up bank teller operation referred to in Problems 1 to determine the following operating characteristics for the system: a.) The probability that no customers are in the system. b.) The average number of customers waiting. c.) The average number of customers in the system. d.) The average time a customer spends waiting. e.) The average time a customer spends in the system. f.) The probability that arriving customers will have to wait for service.

Explanation:

Arrival rate \lambda = 24 customers per hour or 0.4 customers per minute

Service rate \mu = 36 customers per hour or 0.6 customers per minute (from problem 1)

a.) The probability that no customers are in the system , P0 = 1 - \lambda / \mu

= 1 - (24/36) = 1/3 = 0.3333

b.) The average number of customers waiting

Lq = \lambda^2 / [\mu(\mu - \lambda)] = 242 / [36 * (36 - 24)] = 1.33

c.) The average number of customers in the system.

L = Lq + \lambda / \mu = 1.33 + (24/36) = 2

d.) The average time a customer spends waiting.

Wq = \lambda / [\mu(\mu - \lambda)] = 24 / [36 * (36 - 24)] = 0.0555 hr = 3.33 min

e.) The average time a customer spends in the system

W = Wq + 1/\mu = 0.0555 + (1/36) = 0.0833 hr = 5 min

f.) The probability that arriving customers will have to wait for service.

= 1 - P0 = 1 - 0.3333 = 0.6667

3 0

1 year ago

You are working on a documentation file userNotes.txt with some members of your software development team. Just before the file

neonofarm [45]

Answer: Provided in the explanation section

Explanation:

The question says :

You are working on a documentation file userNotes.txt with some members of your software development team. Just before the file needs to be submitted, you manager tells you that a company standard requires that one blank space be left between the end-of-sentence period (.) and the start of the next sentence. (This rule does not affect sentences at the very end of a line.) For example, this is OK: Turn the knob. Push the "on" button. This is not: Turn the knob. Push the "on" button. Asking around among your team members, you discover that some of them have already typed their sentences in the approved manner, but others have inserted two or even more blanks between sentences. You need to fix this fast, and the first thing you want to do is to find out how bad the problem is. What command would you give to list all lines of userNotes.txt that contain sentence endings with two or more blanks before the start of the next sentence?

Solution:

Here, our fundamental aim is to look for the content which is having single space between different sentences. As it sentences finishing with . Going before with single and various spaces we have to channel and match just e the sentences which are finishing with ". "

For this we use order called "GREP" in Unix. "GREP " represents worldwide quest for standard articulation. This order is utilized inquiry and print the lines that are coordinating with determined string or an example. The example or indicated string that we have to look through we call it as customary articulation.

Syntax of GREP:

GREP [OPTIONS] PATTERN [FILE_NAME]

For our solution please follow the command

GREP "*\•\s$" userNotes.txt

Now it will display a all the lines having . Followed by single space.

6 0

1 year ago

What is printed by the following program provided all necessary standard header files are included? Explain each line of the out

Ostrovityanka [42]

<u>Output:</u>

f1 in A

f2 in A

f1 in B

f2 in A

f1 in A

f2 in A

f1 in B

f2 in B

<u>Explanation:</u>

In this snippet, the code makes use of virtual functions. A virtual function is defined as a function that is defined in the base class and redefined in the derived class. If the derived function accesses the virtual function, the program will get executed with the derived class’s version of the function.

In this code, we define the virtual function f1() in class A and also redefine it in class B which is the derived class of A. While executing the program, the function g which takes the object b (class B’s object) as a parameter. It will print class B’s version of f1() rather than class A’s version. This is working off the virtual function.

8 0

1 year ago

You buy a $3,500 car and finance it through the car dealer. the contract says if you are two months delinquent with your payment

photoshop1234 [79]

The car owner acquired his car because the car dealer offers him a financing credit services. It is part of the contract that the car owner should pay in monthly basis. It is a fact and part of the contract also that once the car owner failed to pay 2 months of his contribution, the car dealer will get his car back.

6 0

2 years ago

Compare the elements of the basic Software Development Life Cycle with 2 other models. How are they similar? How are they differ

Artemon [7]

Answer:

Explanation:

One of the basic notions of the software development process is SDLC models which stands for Software Development Life Cycle models. SDLC – is a continuous process, which starts from the moment, when it’s made a decision to launch the project, and it ends at the moment of its full remove from the exploitation. There is no one single SDLC model. They are divided into main groups, each with its features and weaknesses. The most used, popular and important SDLC models are given below:

1. Waterfall model

2. Iterative model

3. Spiral model

4. V-shaped model

5. Agile model

Stage 1. Planning and requirement analysis

Each software development life cycle model starts with the analysis, in which the stakeholders of the process discuss the requirements for the final product.

Stage 2. Designing project architecture

At the second phase of the software development life cycle, the developers are actually designing the architecture. All the different technical questions that may appear on this stage are discussed by all the stakeholders, including the customer.

Stage 3. Development and programming

After the requirements approved, the process goes to the next stage – actual development. Programmers start here with the source code writing while keeping in mind previously defined requirements. The programming by itself assumes four stages

• Algorithm development

• Source code writing

• Compilation

• Testing and debugging

Stage 4. Testing

The testing phase includes the debugging process. All the code flaws missed during the development are detected here, documented, and passed back to the developers to fix.

Stage 5. Deployment

When the program is finalized and has no critical issues – it is time to launch it for the end users.

SDLC MODELS

Waterfall – is a cascade SDLC model, in which development process looks like the flow, moving step by step through the phases of analysis, projecting, realization, testing, implementation, and support. This SDLC model includes gradual execution of every stage completely. This process is strictly documented and predefined with features expected to every phase of this software development life cycle model.

ADVANTAGES

Simple to use and understand

DISADVANTAGES

The software is ready only after the last stage is over

ADVANTAGES

Management simplicity thanks to its rigidity: every phase has a defined result and process review

DISADVANTAGES

High risks and uncertainty

Iterative SDLC Model

The Iterative SDLC model does not need the full list of requirements before the project starts. The development process may start with the requirements to the functional part, which can be expanded later.

ADVANTAGES

Some functions can be quickly be developed at the beginning of the development lifecycle

DISADVANTAGES

Iterative model requires more resources than the waterfall model

The paralleled development can be applied Constant management is required

Spiral SDLC Model

Spiral model – is SDLC model, which combines architecture and prototyping by stages. It is a combination of the Iterative and Waterfall SDLC models with the significant accent on the risk analysis.

4 0

1 year ago