Answer:
- make a promise
- write it in a diary
- make a scrap book
- create a video
- make a picture book
I hope this helps
<h2>I know you will do great!!!</h2>
Answer:
The answer to this question can be described as follows:
a) True
b) False
c) False
d) False
e) False
f) True
Explanation:
In option a, PATA device is used to connect the hard drives into systems. Its title suggests, depends on parallel transmitting technology, contrasted to the compulsive-signaling tools.
In option b, SATA1 is a first-generation device, and SATA3 third generation, that's why it is wrong.
In option c and option d, both were wrong because RAID 0 and RAID 5 are Cloud computing software for storing data.
In option e, An internal SATA data cable with an eSATA port can not be used as an additional connection is provided by the eSATA port.
In option f, It is true because there are 7 pins in internal SATA.
Answer:
Option 3) Boolean is the correct answer
Explanation:
Let us see all the data types given in the options.
Alphabetic datatype stores alphabets and characters.
Float stores numbers with decimal points.
Boolean is a binary data type that can only have two values either one and zero or true and false.
Integer stores positive and negative numbers.
Looking at all the definitions we can conclude that Boolean will be the suitable variable to store the answers as the answer can only have one of the two values from true and false.
Hence,
Option 3) Boolean is the correct answer
Answer:
The solution code is written in Python
- def findSmallest(vec, start):
-
- index = start
- smallest = vec[start]
-
- for i in range(start + 1, len(vec)):
- if(smallest > vec[i]):
- smallest = vec[i]
- index = i
-
- return index
Explanation:
Firstly we can define a function findSmallest() that takes two input parameters, a vector, <em>vec</em>, and a starting position, <em>start </em> (Line 1).
Next, create two variables, <em>index</em> and <em>smallest</em>, to hold the current index and current value where the smallest number is found in the vector. Let's initialize them with <em>start</em> position and the value held in the<em> start </em>position (Line 3-4).
Next, create a for-loop to traverse through the next value of the vector after start position and compare it with current <em>smallest </em>number. If current <em>smallest</em> is bigger than any next value in the vector, the <em>smallest </em>variable will be updated with the new found lower value in the vector and the index where the lower value is found will be assigned to variable<em> index</em>.
At the end return index as output.
