3 years ago

Which action demonstrates good workplace etiquette?

Business

1 answer:

Triss [41]3 years ago

3 0

Answer:

A is the answer of the question

You might be interested in

High income countries with larger governments as a share of gdp have generally

Scorpion4ik [409]

Answer: High income countries with larger governments as a share of GDP have generally grown at a slower rate than the countries with smaller governments.

Explanation: Developing countries or countries with less money typically grow at a faster rate than higher income countries because returns related to capital are not as strong. In richer countries, they have higher capital and tend to grow at a slower rate.

5 0

3 years ago

The method of slicing digital messages into parcels, transmitting them along different communication paths, and reassembling the

Nana76 [90]

Is called packet switching

5 0

3 years ago

James has a marginal tax rate of 24%. He suddenly realizes that he neglected to include a $7,000 tax deduction. How will this ov

Alex17521 [72]

Answer:

hmhgdmthde

Explanation:

7 0

3 years ago

A regulatory agency that protects workers is

bagirrra123 [75]

Answer:

answer is A

Explanation:

6 0

3 years ago

) A company determines that its marginal revenue per day is given by R'(t) = 100et , R(0) = 0, where R(t) = the revenue, in doll

marysya [2.9K]

Answer:

The answer is below

Explanation:

The marginal revenue R'(t) = $100e^t$ and the marginal cost C'(t) = 140 - 0.3t.

The total profit is the difference between the total revenue and total cost of a product, it is given by:

Profit = Revenue - Cost

P(T) = R(T) - C(T)

P(T) = ∫ R'(T) - C'(T)

Hence the total profit from 0 to 5 days is given as

$P(T) = \int\limits^0_5 {(R'(T)-C'(T))} \, dt= \int\limits^0_5 {(100e^t-(140-0.3t))} \, dt\\ \\P(T)= \int\limits^0_5 {(100e^t-140+0.3t))} \, dt\\\\P(T)= \int\limits^0_5 {100e^t} \, dt- \int\limits^0_5 {140} \, dt+ \int\limits^0_5 {0.3t} \, dt\\\\P(T)=100\int\limits^0_5 {e^t} \, dt- 140\int\limits^0_5 {1} \, dt+0.3 \int\limits^0_5 {t} \, dt\\\\P(T)=100[e^t]_0^5-140[t]_0^5+0.3[\frac{t^2}{2} ]_0^5\\\\P(T)=100(147.41)-140(5)+0.3(12.5)=14741-700+3.75\\\\P(T)=14045$

3 0

4 years ago