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padilas [110]
1 year ago
13

Merck, a pharmaceutical company, has taken thousands of drugs through the federal approval process and so can do it more cost ef

ficiently than many competitors who are new to the industry. These cost savings that come to Merck in the drug approval process are an example of:_______
Business
1 answer:
melomori [17]1 year ago
4 0

Answer:

learning effects

Explanation:

Learning effects: In economics, the term "learning effects" is described as the process through which specific education is considered as increasing productivity and therefore results in producing higher wages. It gives an insight to the company to develop some competitive advantage by decreasing some of the production costs. However, the employees are focused on working more efficiently, decrease in the number of wastes and defects on several products.

In the question above, the given statement signifies the leaning effects.  

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As the VP of Global Marketing, what business objective do you want Holden Evan to achieve in Tuatara? Any choice will bring its
Vlada [557]

Answer:

The business objective that I want Holden Evan to achieve in Tuatara is to sell their products throughout the Tuatara territory.

Explanation:

As the VP of Global Marketing, the business objective that i want Holden Evan to achieve in Tuatara is to sell their products throughout the Tuatara territory reason been that Holden Evan is a multinational corporation that deal in selling of beauty products as well as other consumer goods and since Tuatara is an emerging market for consumer products, this means that Holden Evan’s main aim and objective in Tuatara territory should be to manufacture and sell their products throughout the Tuatara territory.

8 0
2 years ago
g Tadeo Corp. has provided a part of its budget for the second​ quarter: Apr May Jun Cash collections $ 42 comma 000 $ 45 comma
timofeeve [1]

Answer:

A. 68,800

Explanation:

Cash balance at end of April is = Beginning cash balance on April 1st + Cash collection in April - Purchase of Materials in APril - Operating Expense in April - Capital Expenditures in APril =  14000 + 42000 - 7000 - 7000 - 5000 = 37000

Cash balance at end of May is = Beginning cash balance in May + Cash collection in May - Purchase of Materials in May - Operating Expense in May = 37000 + 45000 - 7200 - 6000 = 68,800

4 0
1 year ago
Hybrid cars are touted as a "green" alternative; however,the financial aspects of hybrid ownership are not as clear. Consider th
lord [1]

Answer:

a)

the hybrid model initially costs $5,200 more than the regular model, plus you have another $330 in extra ownership costs per year. If you plan to own the hybrid car for 6 years, then you must recoup $5,200 / 6 = $866.67 + $330 = $1,196.67 per year.

the cost of driving 1 mile with the hybrid car = $3.60 / 27 = $0.1333

the cost of driving 1 mile with the regular model = $3.60 / 19 = $0.1895

you will save = $0.0562 per mile driven

you would need to drive $1,196.67 / $0.0562 = 21,293 miles per year to make the decision worth it

b)

if you only drive 15,500 miles per year, then you would need to save $0.0772 per mile

that would only result if gasoline's price was:

x/19 - x/27 = 0.0772

0.0526x - 0.037x = 0.0772

0.0156x = 0.0772

x = 0.0772 / 0.0156 = $4.95 per gallon

c)

you must first determine the present value of all additional expenses related to purchasing a hybrid:

year         cash flow

0                -5,200

1                 -330

2                -330

3                -330

4                -330

5                -330

6                -330

Using a financial calculator, the PV = -$6,637.24

now we must use an annuity formula to determine the annual savings required using a 10% discount rate and 6 periods:

annual savings = $6,637.24 / 4.3553 (PV annuity factor, 10%,  6 periods) = $1,523.95

so you must save $1,523.95 per year and that is equivalent to $1,523.95 / $0.0562 = 27,116.47 = 27,116 miles

d)

you also need to save $1,523.95, but you only drive 15,500 miles, so the savings per mile = $0.0983

x/19 - x/27 = 0.0983

0.0526x - 0.037x = 0.0983

0.0156x = 0.0983

x = 0.0983 / 0.0156 = $6.30 per gallon

5 0
1 year ago
Calculate the fair present values of the following bonds, all of which pay interest semiannually, have a face value of $1,000, h
Mila [183]

Answer:

the bonds' current market value = PV of face value + PV of coupon payments

a. The bond has a 6 percent coupon rate.

PV of face value = $1,000 / (1 + 5%)²⁴ = $310.07

PV of coupon payments = 30 x 13.799 (PV annuity factor, 5%, 24 periods) = $413.97

bond's market value = $724.04

b. The bond has a 8 percent coupon rate.

PV of face value = $1,000 / (1 + 5%)²⁴ = $310.07

PV of coupon payments = 40 x 13.799 (PV annuity factor, 5%, 24 periods) = $551.96

bond's market value = $862.03

3 0
1 year ago
A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is kno
raketka [301]

Here is the correct question.

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to \dfrac{1}{0.85} due to the stress that causes the failure (strength)  is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load  must be reduced to \dfrac{1}{1.1} because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to \dfrac{1}{0.95}  because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

\sigma' = \dfrac{P'}{A'}

where,

\sigma = stress

P = load

A = cross-sectional area of the cylinder

∴

\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}

replacing P' with \dfrac{1}{1.1}P   and d' with \dfrac{1}{0.95}d, we have:

\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }

\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }

\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }

\sigma' =\sigma \times 0.82045

\dfrac{\sigma' }{\sigma } =0.82045

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

n_d = \dfrac{loss  \ of  \ function \  parameter }{maximum \  allowable \ parameter}

where;

the design factor = n_d

n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}

the design factor  n_d = 1.434.

Thus,  the minimum design factor that will guarantee that the part will not fail. = 1.434

7 0
1 year ago
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