Answer:
=IF(D15<1500, "Discontinue", "No Change")
Explanation:
In an Excel environment, you should navigate to cell F15 and apply the IF function. The structure of the IF function is
IF(<condition>, <value if true>, <value if false>)
Your condition is placed before the first comma, which is D15 < 1500.
The second segment, located before the second comma, is the text to display when D15 is below 1500, which is "Discontinue".
Finally, the last part should show text if D15 is 1500 or higher, which is "No Change".
The complete formula can be expressed as =IF(D15<1500, "Discontinue", "No Change")
Response: explained in the explanation section
Explanation:
Given that:
Assume D(k) =║ true if [1::: k] is a valid sequence of words, or false otherwise
the sub problem s[1::: k] is a valid sequence of words IFF s[1::: 1] is valid and s[ 1 + 1::: k] is a valid word.
Thus, we derive that D(k) is defined by the following recurrence relation:
D(k) = ║ false max(d[l] ∧ DICT(s[1 + 1::: k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1::: k] ∦ array of boolean variables.
for a ← 1 to m
do;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1::: a])
d (a) ← True
(b). Algorithm Output
if D[k] == True
stack = temp stack ∦stack assists in displaying the strings in order
c = k
while C > 0
stack push (s [w(c)]::: C] // w(p) denotes the index in s[1::: k] of the valid word // at position c
P = W (p) - 1
output stack
= 0 =
cheers, I hope this aids you!!!
The data for the middle node shows as –250.... Please outline the pseudocode to substitute the middle node in the linked list. Assume that the head pointer is designated as Head_ptr and the entry data for the new node is defined as Entry.
Answer:
Below is the Python code with suitable comments.
Explanation:
#Input file name acquisition
filename=input('Enter the input file name: ')
#Opening the input file
inputFile = open(filename,"r+")
#Dictionary definition.
list={}
#Read and split file content using a loop
for word in inputFile.read().split():
#Check if the word exists in the file.
if word not in list:
list[word] = 1
#Increment count by 1
else:
list[word] += 1
#Closing the file.
inputFile.close();
#Output a blank line
print();
#Sorting words according to their ASCII values.
for i in sorted(list):
#Display unique words along with their
#frequencies in alphabetical order.
print("{0} {1} ".format(i, list[i]));
Answer:
Given Data:
myBytes BYTE 10h, 20h, 30h, 40h
myWords WORD 3 DUP(?), 2000h
myString BYTE "ABCDE"
From the supplied information, we can derive that:
(a). a. EAX = 1
b. EAX = 4
c. EAX = 4
d. EAX = 2
e. EAX = 4
f. EAX = 8
g. EAX = 5
