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Ganezh [65]
2 years ago
6

Mr. Cooper would like to customize his Excel software so his students can create an electronic graph in Excel for their lab repo

rts. Which process would best help the students locate the chart tools options?
creating a graph template
adding the Chart Tools option as a main tab
adding "Create Chart" to the Quick Access toolbar
adding "Graph Tools" as an add-in in the Options dialog box
Computers and Technology
1 answer:
Neko [114]2 years ago
5 0
The third one probably
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Modern operating system decouple a process address space from the machine's physical memory. list two advantages of this design.
likoan [24]

The two advantages of modern operating systems is that it has the capacity in helping programs to be loaded faster compared to other older versions and it also has the ability of having to run programs that are large enough in which is in the main memory even if it is smaller compared to the size of its program.

4 0
2 years ago
A computer has 4 GB of RAM of which the operating system occupies 512 MB. The processes are all 256 MB (for simplicity) and have
Afina-wow [57]

Answer:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

Explanation:

Previous concepts

Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"

Solution to the problem

For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:

Available = 4096 MB - 512 MB = 3584 MB

Now we can find the maximum simultaneous process than can use with this:

\frac{3584 MB}{256 MB/proc}= 14 processes

And then we can find the maximum wait I/O that can be tolerated with the following formula:

1- p^{14}= rate

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

3 0
2 years ago
Consider the following skeletal C program: void fun1(void); /* prototype */ void fun2(void); /* prototype */ void fun3(void); /*
natita [175]

Answer:

Check the explanation

Explanation:

a) main calls fun1; fun1 calls fun2; fun2 calls fun3

fun3()                                        d, e, f

fun2()                                        c, d, e

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "c" of fun2() will be visible

   variable "b" of fun1() will be visible

   variable "a" of main() will be visible

b) main calls fun1; fun1 calls fun3

fun3()                                        d, e, f

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "b, c" of fun1() will be visible

   variable "a" of main() will be visible

c) main calls fun2; fun2 calls fun3; fun3 calls fun1

fun1()                                        b, c, d

fun3()                                        d, e, f

fun2()                                        c, d, e

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun1().

   In fun1(), the local variables "b, c, d" of fun1() will be visible

   variable "e, f" of fun3() will be visible

   variable "a" of main() will be visible

d) main calls fun1; fun1 calls fun3; fun3 calls fun2

fun2()                                        c, d, e

fun3()                                        d, e, f

fun1()                                        b, c, d,

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun2().

   In fun2(), the local variables "c, d, e" of fun2() will be visible

   variable "f" of fun3() will be visible

     variable "b" of fun1() will be visible

   variable "a" of main() will be visible

The last function called will comprise of all its local variables and the variables other than its local variables from all its preceding function calls till the main function.

8 0
2 years ago
Nancy would like to configure an automatic response for all emails received while she is out of the office tomorrow, during busi
Stella [2.4K]

she can appoints someone she trust to act on her behalf

4 0
2 years ago
Read 2 more answers
Clara writes online articles based on world religions. Her articles have references to dates. Which HTML element will help her d
brilliants [131]

Answer:

Available Options are :

A <b>

B <l>

C <body>

D <small>

Ans : A <b>

Explanation:

Normally AD and BC are written in bold letters.

for example : The terms anno Domini (AD) and before Christ (BC).

Within available option the best option to choose is <b>

The <body> element used to define the document body.

The <small> tag used to defines smaller text

But, the best option to choose is in case if they are using the HTML5  then html <time> element will be used to display BC date and AD dates.

for ex : <time datetime="-314-07-01"

calendar="Ancient Roman">1 July 314 BC</time>

3 0
2 years ago
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