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zubka84 [21]
2 years ago
5

Alex is nearing graduation. His counselor showed him this information. Alex's College Costs & Payment Options per Year Costs

Methods of Payment Tuition & Fees Grants & Scholarship $10,100 $12,500 Room & Board Work-Study $11,750 $8,000 If his parents cannot help Alex with college, and two of his scholarships will be awarded to other students if he does not accept them immediately, which is the best option for him? He could work each summer to earn the $1,350 difference. He could wait a year for college and earn the difference. He could abandon his plans for college because he does not have enough money. He could take out a student loan for the difference of $1,350, plus an extra $2,000 a year for mad money.
Computers and Technology
2 answers:
Naya [18.7K]2 years ago
7 0

Answer:

The correct option is option A.

Explanation:

As the options are mentioned in the question but are not discernable clearly, thus the options are as below:

a. He could work each summer to earn the $1,350 difference.

b. He could wait a year for college and earn the difference.

c. He could abandon his plans for college because he does not have enough money.

d. He could take out a student loan for the difference of $1,350, plus an extra $2,000 a year for mad money.

As per the minimum wage of 9 dollars per hour and a 40 hour week, Alex can earn the difference in roughly 4 weeks.

The correct option is A as it is the most logical option available.

Option b is not correct as with waiting one year he will fall behind his peers such that the the amount is also not significantly large.

Option c is not correct as the amount required is not significant and means are available to easily cover up the difference.

Option d is not correct as with the student loan, he has to return the amount with a higher interest rate.

Pie2 years ago
6 0

Answer:

A

Explanation:

In my opinion, it is best he works the extra $1300 which wouldn't require him to abandon his college plans, wait an extra year or get himself into the trouble of loans.

Cheers

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An encryption system works by shifting the binary value for a letter one place to the left. "A" then becomes: 1 1 0 0 0 0 1 0 Th
Vika [28.1K]

Answer:

The hexadecimal equivalent of the encrypted A is C2

Explanation:

Given

Encrypted binary digit of A = 11000010

Required

Hexadecimal equivalent of the encrypted binary digit.

We start by grouping 11000010 in 4 bits

This is as follows;

1100 0010

The we write down the hexadecimal equivalent of each groupings

1100 is equivalent to 12 in hexadecimal

So, 1100 = 12 = C

0010 is represented by 2 in hexadecimal

So, 0010 = 2

Writing this result together; this gives

1100 0010 = C2

Going through the conversion process;

A is first converted to binary digits by shifting a point to the left

A => 11000010

11000010 is then converted to hexadecimal

11000010 = C2

Conclusively, the hexadecimal equivalent of the encrypted A is C2

8 0
2 years ago
Jason is working on a web page that includes Q&A interactions. Which option should Jason select to engage users in the inter
Vaselesa [24]
E should be the correct answer
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2 years ago
Read 2 more answers
2.32 LAB: Musical note frequencies On a piano, a key has a frequency, say f0. Each higher key (black or white) has a frequency o
Maurinko [17]

Answer:

#include<iostream>

#include<iomanip>

#include<cmath>

using namespace std;

int main()

{

float f0;

cout<<"Initial Frequency: ";

cin>>f0;

for(int n = 1; n<=5;n++)

{

 cout<<f0<<setprecision(2);

 f0 = f0 * pow(2.0,1.0/12);

 cout<<" ";

}

 

return 0;

}

Explanation:

The programming language is not stated; However, one can easily deduce that the program is to be written using C++

The following libraries enable the program to make use of some built in functions

<em>#include<iostream> </em>

<em>#include<iomanip> </em>

<em>#include<cmath> </em>

using namespace std;

int main()

{

This line declares the initial frequency as float

float f0;

This line prompts user for input

cout<<"Initial Frequency: ";

This line gets user input

cin>>f0;

The following iteration calculates and prints the frequency for thenexr 4 keys

<em>for(int n = 1; n<=5;n++) </em>

<em> { </em>

<em>  cout<<f0<<setprecision(2); </em>

<em>  f0 = f0 * pow(2.0,1.0/12); </em>

<em>  cout<<" "; </em>

<em> }  </em>

return 0;

}

7 0
2 years ago
Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS
BlackZzzverrR [31]

Answer:

hello your question lacks the C segment so here is the C segment

while ( k<n )

{v[k] = v[k+1];

     k = k+1; }

Answer : while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

Explanation:

The MIPS assembly code corresponding to the C segment is

while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

4 0
2 years ago
Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color's value can range from 0 to 255
maxonik [38]

Answer:

Follows are the code to this question:

#include <iostream>//defining a header file

using namespace std; //using namespace

int main() //defining main method

{

int red,green,blue,s; //defining integer variable

cout<<"Enter value: \n ";//print message

cin>>red>>green>>blue; //input value

if(red<green && red<blue)//defining if block that checks red value

s=red;//store red variable value to s variable

else if(green<blue)//defining else if block that checks green value less then blue

s=green;//store green variable value in s variable

else//defining else block

s=blue; //store blue variable value in s variable

//calculating red, green, blue value

red=red-s;//store red value

green=green-s;//store green value

blue=blue-s;//store blue value

cout<<red<<" "<<green<<" "<<blue;

}

Output:

Enter value:

130  

50

130

80 0 80

Explanation:

In the above code, inside the Main method, four integer variable "red, green, blue, and s" is defined, in which "red, green, and blue" is used for input the value from the user end.

  • In the next step, a conditional statement is used, that checks the red variable value, if the condition is true, it will store its value in the "s" variable, otherwise, it will go to else if block.
  • In this block, the green variable checks its value less than then blue variable value, if the condition is true, it will store the green value in the "s" variable, otherwise, it will goto else block.
  • In this block, it will store the blue variable value in the "s" variable, and subtract the value of "red, green, and blue" value from "s" and store its value, and at the last, it will print its value.    
4 0
2 years ago
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