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marshall27 [118]

3 years ago

11

Store operations has installed a new inventory ordering system that is expected to reduce the time it takes to place orders by 2

0%

Computers and Technology

1 answer:

nikklg [1K]3 years ago

4 0

Answer:

THIS OPERATION IS ALSO a machine called the chef bot or order taker

Explanation:

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rewona [7]

Answer:

WHERE ARE YOU FROM

Explanation:

TRANSLATE IT TO ENGLISH

5 0

3 years ago

Why is it important to send a cover letter with a resume?

Novay_Z [31]

It's important to do so because you want to be sure it looks professional, just like it's important to add a cover to a report. It's also the first look at the resume and gives the overall look of it. It lets the job or employer know the specifics and into the resume talk more of what you do. It's like the overall information page..... Hope this helps.

5 0

3 years ago

Read 2 more answers

1.Write the Qbasic program to find sum of any 10 different numbers.

dybincka [34]

Answer:

1.

DIM myArray(10) as INTEGER

LET A = 0

FOR I = 1 TO 10 STEP 2

INPUT “INPUT NUMBER”; myArray(i)

LET A = A + myArray(i)

NEXT

PRINT A

END

2.

REM PROGRAM FOR CALCULATING THE SIMPLE INTEREST

CLS

INPUT “INPUT THE PRINCIPAL”; P

INPUT “INPUT THE TIME”; T

INPUT “INPUT THE RATE”;R

SI = P* T * R / 100

PRINT “SIMPLE INTEREST =”; SI

END

Explanation:

Please find the respective programs in the answer section.

4 0

3 years ago

Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a

kirill115 [55]

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = $\frac{CPI*Instruction Count}{ClockRate}$

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is $=\frac{2.5*100,000}{750*10^6}$

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = = $\frac{2.5*Instruction}{750*10^6}$

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

3 0

3 years ago

Which kind of device is hardware capable of transferring items from computers and devices to transmission media and vice versa?

Mekhanik [1.2K]

Would it be a flash drive?

7 0

3 years ago