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baherus [9]
1 year ago
5

your monthly living expenses are $1500 on an income of $1,650 per month. your goal is to have an emergency fund of 4 times your

monthly living expenses. your emergency fund savings account has $2,400 and you put the remainder of your monthly income into the emergency fund each month. how much more money would you have to save each month to complete your emergency fund in 12 months?
Computers and Technology
1 answer:
Varvara68 [4.7K]1 year ago
8 0
You will need 6000$ in your Ef. You already got 2400 so now u need 3600$. 3600/12= 300$ , which is the amount of money you need to save, you are already saving 150$ so now you need to save 150 more.
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A(n) _____ is the highest educational degree available at a community college. master bachelor associate specialist
baherus [9]
<span>An associate's degree requires two years of academic study and is the highest degree available at a community college</span>
5 0
2 years ago
Read 2 more answers
Assume the availability of class named IMath that provides a static method, toThePowerOf which accepts two int arguments and ret
geniusboy [140]

Answer:

cubeVolume = IMath.toThePowerOf(cubeSide, 3);

Explanation:

Following is the explanation for above statement:

Left side:

cubeVolume is a variable with data-type int, it will store the integer value that is the output from right side.

Right side:

  • IMath is the class name.
  • toThePowerOf is the built-in function that takes two arguments of data type int. First is the base and second is the power(exponent) separated by comma. In place of first argument that is the base variable we will pass the variable cubeSide that has been declared and initialize.
  • Now the output will be stored in the variable cubeVolume.

i hope it will help you!

4 0
2 years ago
Suppose a computer has 16-bit instructions. The instruction set consists of 32 different operations. All instructions have an op
Kazeer [188]

Answer:

2^7= 128

Explanation:

An instruction format characterizes the diverse part of a guidance. The fundamental segments of an instruction are opcode and operands. Here are the various terms identified with guidance design:  Instruction set size tells the absolute number of guidelines characterized in the processor.  Opcode size is the quantity of bits involved by the opcode which is determined by taking log of guidance set size.  Operand size is the quantity of bits involved by the operand.  Guidance size is determined as total of bits involved by opcode and operands.

6 0
2 years ago
Which of the following is a collection of unprocessed items, which can include text, numbers, images, audio, and video?A. DataB.
Tema [17]

Answer:

Option A is the correct answer for the above question.

Explanation:

Data can be defined as raw fact which can be useful when it will be processed. The processed data can be formed as information. The data can be anything. It can b e text or audio or images or video. The above question asked about the term which is an unprocessed item and can form information after processing. Then the answer is Data which stated from the option A. So Option A is the correct answer while the other is not because--

  • Option B states about instruction which is useful to process the data.
  • Option C states about Programs that can be formed when one or more instruction is grouped.
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6 0
2 years ago
In the simulation, player 2 will always play according to the same strategy. The number of coins player 2 spends is based on wha
baherus [9]

The simulation, player 2 will always play according to the same strategy.

Method getPlayer2Move below is completed by assigning the correct value to result to be returned.

Explanation:

  • You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.

#include <bits/stdc++.h>  

using namespace std;

bool getplayer2move(int x, int y, int n)  

{

   int dp[n + 1];  

   dp[0] = false;  

   dp[1] = true;  

   for (int i = 2; i <= n; i++) {  

       if (i - 1 >= 0 and !dp[i - 1])  

           dp[i] = true;  

       else if (i - x >= 0 and !dp[i - x])  

           dp[i] = true;  

       else if (i - y >= 0 and !dp[i - y])  

           dp[i] = true;  

       else

           dp[i] = false;  

   }  

   return dp[n];  

}  

int main()  

{  

   int x = 3, y = 4, n = 5;  

   if (findWinner(x, y, n))  

       cout << 'A';  

   else

       cout << 'B';  

   return 0;  

}

8 0
2 years ago
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