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3 years ago

In an oogilyboogily market, consumer choice is?

Business

1 answer:

Zielflug [23.3K]3 years ago

4 0

Answer:

limited.

Explanation:

In an oligopoly market, few firms dominate a market with many buyers. The firms may sell similar or differentiated products. The oligopoly stature is characterized by heavy advertisement to win customers. Each firm sets its price.

Because few firms dominate the market, customers do not have a wide range of variety to select from. For example, the mobile phone business is dominated by 5 to 6 companies. Customers' choices are limited to these service providers.

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You interview with an athletic footwear manufacturer that has annual advertising expenditures of $32 million and total sales rev

son4ous [18]

Answer:

elastic.

Explanation:

The advertising elasticity of demand measures how sensitive a market and sales are to marketing expenses. Advertising elasticity is calculated by dividing the change in quantity demanded by the percentage change in advertising expenses. Generally products with low advertising elasticity tend to have elastic demands.

8 0

3 years ago

Is mary kay an "international firm", a "multinational firm", or a "transnational firm" as defined in this course and based on it

masha68 [24]

Here is the answer. Mary Kay Inc. is a transnational firm that specializes in women’s cosmetics as defined in this course and based on its marketing strategy. Hope this is the answer that you are looking for. Hope this answers your question. Have a great day!

4 0

4 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:

For 1st algorithm: 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%.

For 2nd algorithm: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%.

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

1st Algorithm

/*dgemm0: simple ijk version triple loop

algorithm*/

for (i=0; i<n; i++)

for (j=0; j<n; j++)

for (k=0; k<n; k++)

c[i*n+j] += a[i*n+k] * b[k*n+j];

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

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

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%.

2nd Algorithm

/*dgemm1: simple ijk version triple loop

algorithm with register reuse*/

for (i=0; i<n; i++)

for (j=0; j<n; j++) {

register double r = c[i*n+j];

for (k=0; k<n; k++)

r += a[i*n+k] * b[k*n+j];

c[i*n+j] = r;

}

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

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