Answer:D. Industrial Internet of things.
Explanation: This is the term used to describe the interconnection of sensors,Routers and other devices in order to improve the overall quality, efficiency, speed, amount of data collected,analysed and processed over a given period of time and over a wide range of Networks.
Industrial Internet of things is different from the internet of things basically because of their usage. Industrial Internet of things is used in industries where the volume of data is high.
click through rate is the answer
Answer:
The answer is "Option a".
Explanation:
This design is the breakdown for a structure to smaller components to recognize the textural functionalities. This system analysis the built-in the top-down style, that defines and it does not describe some first-level components, which is often known as a staggered layout, and wrong choices can be described as follows:
- In option b, It is wrong because it is not used in web pages.
- In option c, It does not start with individual commands, that's why it is incorrect.
- In option d, It is wrong because it uses loops and classes, but it can't decompose the problem.
Answer:
To create a console that is suitable for children and families.
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 = 
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is 
= 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 = =
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