Answer:
Hard to change ; No digital skills among staff
Explanation:
Traditional ad / marketing agencies are the agencies promoting brands through offline ways. Eg : Banners, Pamphlets etc
Digital Marketing agencies are agencies promoting through online ways. Eg : E mail marketing, Social media marketing etc.
Digital Marketing needs more technical expertise than traditional, conventional marketing. So, traditional marketers & their staff face adaptability issues in adapting to the new technically upgraded marketing approaches. Such because their team & staff members have low techno - digital skills, are accustomed to conventional marketing practices.
<span>Variances allow the business owner to supervise
their business better by taking well-versed decisions based on how the business
really performed against the budgeted performance. Additionally, it also
highlights reasons or different causes for the disparity in the projected
income or expenses.</span>
Answer:
B
Explanation:
Original Cost -$120,000
Useful life -10 years
Residual Value - $20000
Annual depreciation - $(120,000-20000)/10 = $10,000
Accumulated depreciation for 4 years = 10*4= $40000
Book value at disposal = $120,000-$40000= $80000
Sales value = $35,000
Loss on disposal = $80,000-$35000= $45,000
Answer: largely efficient
Explanation:
The fact that less than half of all equity fund managers beat the market in most years indicate that the stock market is largely efficient.
According to the strong-form hypothesis of the efficient market, when there is an efficient market, all the private and public information would be reflected in the prices of the stock.
Answer:
The answer is below
Explanation:
The marginal revenue R'(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](https://tex.z-dn.net/?f=P%28T%29%20%3D%20%5Cint%5Climits%5E0_5%20%7B%28R%27%28T%29-C%27%28T%29%29%7D%20%5C%2C%20dt%3D%20%5Cint%5Climits%5E0_5%20%7B%28100e%5Et-%28140-0.3t%29%29%7D%20%5C%2C%20dt%5C%5C%20%5C%5CP%28T%29%3D%20%5Cint%5Climits%5E0_5%20%7B%28100e%5Et-140%2B0.3t%29%29%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D%20%5Cint%5Climits%5E0_5%20%7B100e%5Et%7D%20%5C%2C%20dt-%20%5Cint%5Climits%5E0_5%20%7B140%7D%20%5C%2C%20dt%2B%20%5Cint%5Climits%5E0_5%20%7B0.3t%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D100%5Cint%5Climits%5E0_5%20%7Be%5Et%7D%20%5C%2C%20dt-%20140%5Cint%5Climits%5E0_5%20%7B1%7D%20%5C%2C%20dt%2B0.3%20%5Cint%5Climits%5E0_5%20%7Bt%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D100%5Be%5Et%5D_0%5E5-140%5Bt%5D_0%5E5%2B0.3%5B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%5D_0%5E5%5C%5C%5C%5CP%28T%29%3D100%28147.41%29-140%285%29%2B0.3%2812.5%29%3D14741-700%2B3.75%5C%5C%5C%5CP%28T%29%3D14045)
