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AlladinOne [14]
1 year ago
13

A Color class has three int color component instance variables: red, green, and blue. Write a toString method for this class. It

should return a String consisting of the three color components (in the order red, green, blue) within parentheses, separated by commas, with a '#' prefix, e.g. #(125,30,210)
Computers and Technology
1 answer:
ELEN [110]1 year ago
8 0

Answer:

public String toString() {

return "#(" + red + "," + green + "," + blue + ")";

}

Explanation:

The code:

public String toString() {

return "#(" + red + "," + green + "," + blue + ")";

}

Is a tostring java code, and it is so easy to write one, as compare to other programming languages

Writing a tostring method returns a strinb representation of an object in Java. Normally, the toString method returns the name of the object’s class plus its hash code.

This code creates a new colour object; then the result of its toString method is printed to the console.

Tostring method can be override. The default implementation of toString isn’t very useful in most situations. So, one can just override it.

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A student who used a regression model that included indicator variables was upset when receiving only the following output on th
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Answer:

This is a multicolinearity problem and the student should determine the variable(s) that cause(s) the problem and remove it.

Explanation:

This information means that there exists a linear combination between the independent variables. The problem might have developed due to multicolinearity producing almost perfectly linearly dependent columns.

This could also be as a results of single matrix created when the student use an incorrect indicator variables and included an additional indicator column which created linearly dependent columns.

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2 years ago
Read 2 more answers
B.) Define latency and jitter for service flow in a fixed wireless system.
yuradex [85]
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2 years ago
Use the values in the range A7:A8 to extend the list of years to the range A9:A11. 3. Use AutoFill to fill the range A9:H11 with
jok3333 [9.3K]

Answer:

Question is incomplete and it needs a proper XL sheet image and more details. I have assumed and answered this question.

1. Use the values in the range A7:A8 to extend the list of years to the range A9:A11

Ans : <em>Use Ctrl + select A7: A8 and drag till A9: A11 : This will extend the list </em>

<em />

<em>2. </em>Use AutoFill to fill the range A9:H11 with the formatting from the range A7:H8.

Ans : Apply format immediately to the cells A9: H11 below with the help of Auto Fill Handle. Using Auto fill button we can able to a cell format to the cell bellow of active cell by just double clicking on the auto fill button.

3. In cell B8, create a formula without using a function that multiplies the value in cell B7 by the value in cell B6 and then adds the value in B7. Update the cell reference to cell B6, from a relative reference to a mixed reference by making an absolute reference to row 

Ans : =($B7*$B6)+$B7

4. Copy the formula from cell B8 to the range B9:B11 and then copy the formulas from the range B8:B11 to the range C8:H11.

Ans:

1. Select the cell B8 that contains which has formula to be copied

2. Click inside the formula bar to activate it and select the entire formula by using mouse and keyboard.

3. Copy the selected formula and press Esc to deactivate the formula bar.

4. Select the cell  B9 : B11 in which you want the copy of the formula to appear.

5. Most range names refer to absolute cell references. This means that when you copy a formula that uses a range name, the copied formula will use the same range name as the original.

6. Paste the formula by using references.

7. Repeat the steps of 1 to 6 for B8:B11 for range C8:H11.

5. Switch to the Cost Estimates worksheet. In cell A9, create a formula using the AVERAGE function that calculates the average of the values in the range A5:A7, then copy your formula to cell D9.

Ans : =AVERAGE(A5:A7) and Copy it to D9 using answer given in 4.

Explanation:

8 0
1 year 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
1 year ago
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