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baherus [9]
2 years 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 $2,400 and you put the remainder of your monthly income into the emergency fund each month. how much more money would you have to save each month to complete your emergency fund in 12 months?
Computers and Technology
1 answer:
Varvara68 [4.7K]2 years ago
8 0
You will need 6000$ 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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Sarah is creating and formatting a newsletter for her school. Which page layout rules should she consider when doing this?
lorasvet [3.4K]
Hey!

I believe the answer to 1. Should be A. Rule of Thirds.
6 0
2 years ago
Read 2 more answers
Use Excel to develop a regression model for the Consumer Food Database (using the "Excel Databases.xls" file on Blackboard) to p
m_a_m_a [10]

Answer:

Step 1 : Create an Indicator Variable for metro cities using formula mentioned in formula bar.

Step 2: Filter the Data on Metro cities i.e. select only those cities with Metro Indicator 1.

Step 3: Paste this filtered data to a new sheet.

Step 4: Go to Data - Data Analysis - Regression

Step 5: Enter the range of Y-variable and X-variable as shown. Select Output range and click on residuals. It will give you Output Summary and the Predicted Values along with Residuals

Please see attachment

5 0
2 years ago
Given sphereRadius and piVal, compute the volume of a sphere and assign to sphereVolume. Look up the equation online (e.g., http
fenix001 [56]

Answer:

  1. #include <stdio.h>
  2. int main()
  3. {
  4.    const double piVal = 3.14159;
  5.    double sphereVolume = 0.0;
  6.    double sphereRadius = 0.0;
  7.    
  8.    sphereRadius = 1.0;
  9.    sphereVolume = 4.0/ 3.0 * piVal * sphereRadius * sphereRadius * sphereRadius;
  10.    
  11.    
  12.    printf("Sphere volume: %lf\n", sphereVolume);
  13.    return 0;
  14. }

Explanation:

Firstly we can identify the formula to calculate volume of sphere which is

Volume = 4/3 \pi r^{3}

With this formula in mind, we can apply this formula to calculate the volume of sphere in Line 10. This is important to perform floating-point division 4.0/3.0 to ensure the resulting value is a floating value as well. Since we have been given piVal and sphereRadius, we can just multiply the result of floating-point division with piVal and sphereRadius and get the sphereVolume value.

At last, display the sphere volume using printf method (Line 13).

6 0
2 years ago
The OSI security architecture provides a systematic framework for defining security attacks, mechanisms, and services. True or F
Alex Ar [27]

Answer:

True is the correct answer for the above question

Explanation:

  • The OSI security architecture is a security framework that is used to secure the data packets when it is going to transfer on the internet for communication.
  • It is used to define the security rules and mechanism which need to secure the data packets.
  • It mainly focuses on Security services, Security attacks, Security mechanisms.
  • The above question-statement states about the framework which defines the security mechanism to secure from the attacks are known as OSI Security architecture which is the correct statement which is described above. Hence True is the correct answer.
3 0
2 years ago
A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an
yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

                                       CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

                    = 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x  10⁹ x 0.75 / 3 Ghz

                  = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

     Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

                = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that  CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore,   numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x  10⁹ / 0.75  = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall  that:

MIPS = Clock rate / CPI X 10⁶

  So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

        MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

  We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

                  MFLOPS₁ = 1777.6

                  MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0
2 years ago
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