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-Dominant- [34]
4 years ago
5

Fosnight Enterprises prepared the following sales budget: The expected gross profit rate is 30% and the inventory at the end of

February was $10,000. Desired inventory levels at the end of the month are 20% of the next month's cost of goods sold. What is the desired beginning inventory on June 1?
Business
1 answer:
Travka [436]4 years ago
8 0

Answer:

$1,960

Explanation:

Complete Questin:

Fosnight Enterprises prepared the following sales budget:

Month Budgeted Sales

March $6,000

April $13,000

May $12,000

June $14,000

The expected gross profit rate is 30% and the inventory at the end of February was $10,000. Desired inventory levels at the end of the month are 20% of the next month's cost of goods sold. What is the desired beginning inventory on June 1?

Sales = 100% – 30%

Gross Profit = 70%

Cost of Goods Sold (CGS)

Therefore, June Sales= $14,000 × 70%

= 9,800 (CGS) × 20%

= $1,960

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A store has two different coupons that customers can use. One coupon gives the customer $15 off their purchase, and the other co
andrey2020 [161]

Answer:

16.25;

g(f(x)) ;

76 ;

f(g(x))

Explanation:

For 15 off

f(x) = x - 15

For 35% off

g(x) = (1 - 0.35)x = 0.65x

g(x) = 0.65x

A.)

For the $15 off coupon :

f(x) = x - 15

f(x) 40 - 15 = 25

For the 35% coupon :

g(x) = (1-0.35)x

g(x) = 0.65(25)

g(x) = 16.25

B.)

Applying $15 off first, then 35%

Here, g is a function of f(x)

g(f(x))

Here g(x) takes in the result of f(x) ;

For the $140 off coupon :

f(x) = x - 15

f(140) = 140 - 15 = 125

For the 35% coupon :

g(125) = (1-0.35)x

g(124) = 0.65(125) = $81.25

C.)

x = 140

g(x) = 0.65x

g(140) = 0.65(140)

g(140) = 91

f(x) = x - 15

f(91) = 91 - 15

f(91) = 76

D.)

Here, F is a function of g(x)

f(g(x))

f(x) = (0.65*140) - 15

6 0
3 years ago
Women speakers who are nervous tend to wobble on their high heels.
Pachacha [2.7K]

Answer:

Yes?

Explanation:

is this a true or false question

if so I think yes but idk

8 0
3 years ago
A company has a selling price of $1,500 each for its printers. Each printer has a 2 year warranty that covers replacement of def
dlinn [17]

Answer:

$103,680

Explanation:

estimated warrant liablity 3% of unid sold at $144

24,000 x 3% x 144 = $103,680

This will be the expected warranty laiblity for the sales of the period, and also the warranty expense.

warranty expense 103,680

warranty liability               103,680

warranty liability    47,000

   inventory                          47,000

to record warranty services

(we use inventory because the company use replacement part, those par are represented in inventory account)

<u>Warrant liablity account</u>

beginning balance 26,000

warranty expense  103,680

warrant serviced    (47,000)

ending balance       82,680

3 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
A factory owner might decide to manufacture shirts in pakistan instead of the united states because
seraphim [82]
Because it is cheaper to manufacture shirts there
6 0
4 years ago
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