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deff fn [24]
4 years ago
15

Jack Palomo has deposited $2,500 today in an account paying 6 percent interest annually. What would be the simple interest earne

d on this investment in five years? If the account pays compound interest, what will be the interest on interest in five years?A) $750; $95.56 B) $150; $845.56 C) $150; $95.56 D) $95.56; $845.56
Business
1 answer:
andrey2020 [161]4 years ago
6 0

Answer:

the answer is a 6 ÷ 2500 ×5

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Question 6
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Answer:

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8 0
3 years ago
Morganton Company makes one product and it provided the following information to help prepare the master budget:
olga nikolaevna [1]

Answer:

1. What is the accounts receivable balance at the end of July?

  • $931,000

2. If we assume that there is no fixed manufacturing overhead and the variable manufacturing overhead is $10 per direct labor-hour, what is the estimated finished goods inventory balance at the end of July?

  • $235,200

3. If we assume that there is no fixed manufacturing overhead and the variable manufacturing overhead is $10 per direct labor-hour, what is the estimated cost of goods sold and gross margin for July?

  • COGS July = 19,000 x $46 = $874,000
  • gross profit July = $456,000

4. What is the estimated total selling and administrative expense for July?

  • $107,000

5. If we assume that there is no fixed manufacturing overhead and the variable manufacturing overhead is $10 per direct labor-hour, what is the estimated net operating income for July?

  • $349,000

Explanation:

budgeted selling price per unit $70

budgeted unit sales:

June                      July                        August                September

units          $$$      units          $$$     units          $$$   units          $$$

8,800        $616     19,000    $1,330   21,000    $1,470  22,000    $1,540

                 $184.8                  $431.2

                                              $399  (from July) <u>$931</u>

                                                                            $441                     $1,029

                                                                                                         $462

ending finished goods inventory:

June                      July                        August                September

units          $$$      units          $$$     units          $$$   units          $$$

3,800                     4,200                    4,400

variable manufacturing overhead per unit = $10 x 2 = $20

direct materials per unit = $12

direct labor per unit = $24

total cost per unit = $56

total ending goods inventory for July = $46 x 4,200 units = $235,200

Revenue July = 19,000 x $70 = $1,330,000

COGS July = 19,000 x $46 = $874,000

gross profit = $456,000

variable S&A expense = $2.00

fixed S&A expense = $69,000

total S&A expense for July = (19,000 x $2) + $69,000 = $107,000

estimated net operating income July = gross margin - S&A = $456,000 - $107,000 = $349,000

6 0
3 years ago
Assume your computer is able to complete 4 double floating-point operations per cycle when operands are in registers and it take
svp [43]

Answer:

<em>For 1st algorithm</em><em>: The total run time is 200.25 s, while that of total waste time is 200 sec and the percentage of the waste time is 99.8%.</em>

<em>For 2nd algorithm:</em><em>The total run time is 100.35 s, while that of total waste time is 100.1 sec and the percentage of the waste time is 99.75%.</em>

Explanation:

As the complete question is not visible, therefore, the question is searched online and following reference question is obtained.

Following data is given as

Floating point operation time=T/4

Memory Access Time =100T

Frequency =2 GHz

Number of Cycles=1000

<u>1st Algorithm</u>

<em>/*dgemm0: simple ijk version triple loop</em>

<em>algorithm*/</em>

<em>for (i=0; i<n; i++)</em>

<em>for (j=0; j<n; j++)</em>

<em>for (k=0; k<n; k++)</em>

<em>c[i*n+j] += a[i*n+k] * b[k*n+j];</em>

First by rewriting the operation inside the inner loop:

= + ×

Now first A, B and C are loaded into the registers so

Load \,Time=3 \times Memory \,Access \,Time=3 \times 100\, T =300\, T

For 2 floating point computations (addition and multiplication)

Computation\, Time=2 \times Floating\, Time\\Computation\, Time=2 \times \frac{T}{4}\\Computation\, Time=\frac{T}{2}

Finally, to store and repeat the cycle as N^3 times the time is estimated as

Store \,Time=Memory\, Access\, Time=100T

Total Run time is given as

T_{run}=N^3 \times [T_{load}+T_{comp}+T_{store}]\\T_{run}=1000^3 \times [300T+\frac{T}{2}+100T]\\T_{run}=1000^3 \times [400.5T]\\T_{run}=200.25 s

Total Wasted time is given as

T_{waste}=N^3 \times [T_{load}+T_{store}]\\T_{waste}=1000^3 \times [300T+100T]\\T_{waste}=1000^3 \times [400T]\\T_{waste}=200 s

Percentage of Waste time is given as

\%age \, waste=\frac{T_{waste}}{T_{run}}\times 100\\\%age \, waste=\frac{200}{200.25}\times 100\\\%age \, waste=99.8\%

<em>The total run time is 200.25 s, while that of total waste time is 200 sec and the percentage of the waste time is 99.8%.</em>

<u>2nd Algorithm</u>

<em>/*dgemm1: simple ijk version triple loop</em>

<em>algorithm with register reuse*/</em>

<em>for (i=0; i<n; i++)</em>

<em>for (j=0; j<n; j++) {</em>

<em>register double r = c[i*n+j];</em>

<em>for (k=0; k<n; k++)</em>

<em>r += a[i*n+k] * b[k*n+j];</em>

<em>c[i*n+j] = r;</em>

<em>}</em>

Initialize register r with the content of C for N2 Times as given as Initialization\,Time=N^2 \times Memory \,Access \,Time=N^3 \times 100\, T

Time for Loading Operands A and B into registers for N3 Times is given as

Load \,Time=N^3 \times 2 \times Memory \,Access \,Time=N^3\times 2 \times 100\, T =N^3\times 200\, T

For 2 floating point computations (addition and multiplication)

Computation\, Time=N^3 \times\frac{T}{2}

Final Memory update to store result in the register r to the memory for N2 Times

Store \,Time=Memory\, Access\, Time=N^2 \times 100T

Total Run time is given as

T_{run}=N^3 \times [T_{load}+T_{comp}]+N^2 \times [T_{linit}+T_{store}]\\T_{run}=1000^3 \times [200T+\frac{T}{2}]+1000^2 \times [100T+100T]\\T_{run}=1000^3 \times [200.5T]+1000^2 \times [200T]\\T_{run}=100.35 s

Total Wasted time is given as

T_{waste}=N^3 \times [T_{load}]+N^2 \times [T_{init}+T_{store}]\\T_{waste}=1000^3 \times [200]+1000^2 \times [100T+100T]\\T_{waste}=1000^3 \times [200T]+1000^2 \times [200T]\\T_{waste}=100.1 s

Percentage of Waste time is given as

\%age \, waste=\frac{T_{waste}}{T_{run}}\times 100\\\%age \, waste=\frac{100.1}{100.35}\times 100\\\%age \, waste=99.75\%

<em>The total run time is 100.35 s, while that of total waste time is 100.1 sec and the percentage of the waste time is 99.75%.</em>

8 0
3 years ago
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