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Daniel [21]
1 year ago
9

A large gambling company needs to be able to accept high volumes of customer wagers within short timeframes for high-profile spo

rting events. Additionally, strict laws prohibit any gambling activities outside of specially-licensed business zones. What is an example of an effective, elastic Cloud solution that can meet this client's needs?
Computers and Technology
1 answer:
ElenaW [278]1 year ago
5 0

Answer:

a mobile app that only accepts wagers based on the user's location.

Explanation:

One effective Cloud Solution would be a mobile app that only accepts wagers based on the user's location. This app would allow the clients to safely and effectively place any wagers which are sent to the gambling company's cloud servers which would be third-party Cloud services. These services will make sure that the gambling company can handle any number of wages at any given time. The mobile app would also only accept wagers from individuals in specific locations in which the law does not prohibit gambling. Therefore, preventing the gambling company from being liable for any losses.

You might be interested in
Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
Complete the function to replace any period by an exclamation point. Ex: "Hello. I'm Miley. Nice to meet you." becomes:
vodomira [7]

Answer:

Here is the complete function:

void MakeSentenceExcited(char* sentenceText) {  // function that takes the text as parameter and replaces any period by an exclamation point in that text

int size = strlen(sentenceText);  //returns the length of sentenceText string and assigns it to size variable

char * ptr;  // character type pointer ptr

ptr = sentenceText;  // ptr points to the sentenceText string

for (int i=0; i<size; i++){  //iterates through the sentenceText string using i as an index

    if (sentenceText[i]=='.'){  // if the character at i-th index of sentenceText is a period

        sentenceText[i]='!'; } } } //places exclamation mark when it finds a period at i-th index of sentenceText

Explanation:

The program works as follows:

Suppose we have the string:

sentenceText = "Hello. I'm Miley. Nice to meet you."

The MakeSentenceExcited method takes this sentenceText as parameter

int size = strlen(sentenceText) this returns the length of sentenceText

The size of sentenceText is 35 as this string contains 35 characters

size =  35

Then a pointer ptr is declared which is set to point to sentenceText

for (int i=0; i<size; i++) loop works as follows:    

1st iteration:

i=0

i<size is true because i=0 and size = 35 so 0<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[0]=='.')

The first element of sentenceText is H

H is not a period sign so the statement inside if statement does not execute and value of i increments to 1. Now i = 1

2nd iteration:

i=1

i<size is true because i=1 and size = 35 so 1<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[1]=='.')

This is the second element of sentenceText i.e. e

e is not a period sign so the statement inside if statement does not execute and value of i increments to 1. Now i = 2

So at each iteration the if condition checks if the character at i-th index of string sentenceText is a period.

Now lets see a case where the element at i-th index is a period:

6th iteration:

i=5

i<size is true because i=5 and size = 35 so 5<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[5]=='.')

This is the character at 5th index of sentenceText i.e. "."

So the if condition evaluates to true and the statement inside if part executes:

sentenceText[i]='!'; statement becomes:

sentenceText[5]='!'; this means that the character at 5th position of sentenceText string is assigned an exclamation mark.

So from above 6 iterations the result is:

Hello!

This loop continues to execute until all the characters of sentenceText are checked and when the value of i gets greater than or equal to the length of sentenceText then the loop breaks.

The screenshot of the program along with its output is attached.

6 0
1 year ago
When checking an id, a server notices a tear in the lamination covering the id, but the other features appear to be valid. the s
KatRina [158]

Answer:

server must ask for a second ID

Explanation:

Based on the information provided within the question it can be said that in this scenario the server must ask for a second ID. If the individual provides a second ID make sure that it is valid. If the individual cannot provide a second ID service can be denied since the ID may have been tampered with since the lamination is coming off.

7 0
2 years ago
Consider an improved version of the Vigen ere cipher, where instead of using multiple shift ciphers, multiple mono-alphabetic su
anzhelika [568]

Answer:

Kasiski’s method for determining 't' works for Vigenère cipher as well. The only difference is therefore in the second stage of the attack. In the second stage, one needs to build a frequency table for each of the 't' keys, and carry out an attack like on the mono-alphabetic cipher. Given a long enough plaintext, this will work successfully.

Explanation:

Kasiski method is a method of attacking polyalphabetic substitution ciphers such as Vigenère cipher. It is also called Kasiski test or Kasiski examination.

The method involve finding the length of the keyword and then dividing the message into that many simple substitution cryptograms. Frequency analysis could then be used to solve the resulting simple substitution.

6 0
2 years ago
Write a statement that assigns finalResult with the sum of num1 and num2, divided by 3. Ex: If num1 is 4 and num2 is 5, finalRes
olganol [36]

Answer:

Written using Python 3:

<em>num1 = int(input("Enter first value: ")) </em>

<em>num2 = int(input("Enter second value: ")) </em>

<em> </em>

<em>finalResult =(num1+num2) / 3 </em>

<em>print("Answer is: ", finalResult)</em>

<em />

INPUT:

Enter first value: 4

Enter second value: 5

OUTPUT:

Answer is: 3

Explanation:

First step is to declare the variables:

  • num1 and num2

Second is to ask for an input:

  • input("Enter first value: ")

Then we need to convert them into integers by encapsulating the line above inside int(). This is a requirement before you can do any mathematical operations with the given values.

  • num1 = int(input("Enter first value: "))

Next is to calculate:

  • <em>finalResult </em><em>=(num1+num2) / 3 </em>
  • Here we first add the values of num1 and num2, <em>then </em>divide the sum by 3.

And to test if the output is correct, let us print it:

  • <em>print("Answer is: "</em><em>,</em><em> </em><em>finalResult</em><em>)</em>
  • print() - when used for strings/texts, use quotation marks like: "text here"
  • but when used to print variables and numbers (integers, floats, etc), there is no need to enclose them with quotation marks.
  • to concatenate string and an integer (in this case the variable finalResult), simply add a comma right after the string.

<em />

6 0
2 years ago
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