A network design engineer has been asked to design the IP addressing scheme for a customer network. The network will use IP addr

Answer:

input price

set change to one hundred minus price

set quarter to change divided by 25

set change to change mod 25

set dime to change divided by 10

set change to change mod 10

set nickel to change divided by 5

set change to change mod 5

set penny to change

print "You acquired an item for price cents and paid me a dollar. Your change is: "

if quarter is equal to 1

print quarter

else if quarter is greater than 1

print quarter with "quarters"

if dime is equal to 1

print dime

else if dime is greater than 1

print dime with "dimes"

if nickel is equal to 1

print nickel

else if nickel is greater than 1

print nickel with "nickels"

if penny is equal to 1

print penny

else if penny is greater than 1

print penny with "pennies"

Explanation:

A clearer way to express the pseudocode above could involve using alternative names for change as:

input price

set change to one hundred minus price

set quarter to change divided by 25

set quarter_remainder to change mod 25

set dime to quarter_remainder divided by 10

set dimes_remainder to quarter_remainder mod 10

set nickel to dimes_remainder divided by 5

set nickel_remainder to dimes_remainder mod 5

set penny to nickel_remainder

print "You purchased an item for" price"cents and provided a dollar. Your change is: "

if quarter is equal to 1

print quarter "quarter"

else if quarter is greater than 1

print quarter "quarters"

if dime is equal to 1

print dime "dime"

else if dime is greater than 1

print dime "dimes"

if nickel is equal to 1

print nickel "nickel"

else if nickel is greater than 1

print nickel "nickels"

if penny is equal to 1

print penny "penny"

else if penny is greater than 1

print penny "pennies"

Next, I will illustrate the pseudocode through an example:

Let's assume price = 75

Thus, change = 100 - price = 100 - 75 = 25

change = 25

quarter = change/25 = 25/25 = 1

quarter = 1

quarter_remainder = change % 25 = 25%25 = 0

dime = 0/10 = 0

At this point, all additional values are 0.

The following message will be displayed on the screen:

You acquired an item for 75 cents and paid me a dollar. Your change is:

Now the program proceeds to the conditional statement

if quarter == 1

This condition holds true because the value of quarter is 1

print quarter "quarter"

Subsequently, this line will appear on the screen:

1 quarter

Thus, the full output of the pseudocode is:

You acquired an item for 75 cents and paid me a dollar. Your change is:

1 quarter

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

• In determining D(n)

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!!!

Answer:

The code placed where "/*Your solution goes here */" is as follows:

while(numInsects<200) // while loop

       {

            System.out.print(numInsects+" "); // output statement

            numInsects=numInsects*2; // operation to double the value.

       }

Output:

• If the input is 16, the result is 16 32 64 128.
• If the input is 25, the result is 25 50 100.

Explanation:

• The above Java code fills the section designated as "Your solution" and will operate correctly.
• The program accepts a user input and continues to print double the value until it reaches 200 or more.
• The while loop evaluates whether the value remains below 200. If so, the operation proceeds; otherwise, it stops.

The top-level domain (TLD) is the highest level in the domain name system (DNS) hierarchy. In the context of the Internet's DNS, the TLD is operationally positioned within the root zone of the namespace. Generally,.com is regarded as the premier choice, followed closely by.net and.org. For a website tailored for mobile optimization,.com would be the most suitable option.