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2 years ago

A manufacturer that sells _ is most likely to employ personal selling.

Business

1 answer:

bonufazy [111]2 years ago

5 0

Answer:

C. Private jets to people.

Explanation:

Personal selling refers to a strategy in which the sales people meet with the customer to convince him/her to buy the product. This strategy is used when the goods of services are costly or technical. According to this, the answer is that a manufacturer that sells private jets to people is most likely to employ personal selling because it is an specialized and costly product that requires to meet the customer to be able to explain everything and encourage him/her to make the purchase.

The other options are not right because they are cheaper products and don't require the sales people to meet with the customer to be able to sell them.

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Insurance is a financial service that allows a

astra-53 [7]

Insurance is a financial service that allows a consumer to share liability with a company.

The answer is C

4 0

2 years ago

Read 2 more answers

Of all customers purchasing automatic garage door openers, 75% purchase chain-driven model. Let X = the number among the next 15

KIM [24]

Answer:

$P(X=x)=C(15,x)\cdot(0.75)^x\cdot(0.25)^{(15-x)}$

Explanation:

The probability mass function (PMF), or frequency function, is the function that gives the probabilities that a discrete random variable take some values.

In this problem, it is requested the frequency function (PMF) for the number of purchasers, among the next 15, who select a chain-driven model.

Then , you need to find, the function that gives P(X=0), P(X=1), P(X=2), P(X=3), . . . up to P(X=15).

Such as any function, the frequency function can be presented as a formula, as a table, or as a graph.

Note that the statement represents a binomial disbribution in which success is that a customer select a chain-driven model and the fail is that a cusotmer does not select a chain-driven model.

The binomial probability for X = the number among the 15 purchasers who select the chain-driven model is given by the formula:

$P(X=x)=C(n,x)\cdot(p)^x\cdot(1-p)^{(n-x)}$

Where:

$C(n,x)=\dfrac{n!}{x!(n-x)!}$

n is the number of times the experiment is performed: 15 in our problem

p is the probability of succes: 0.75 in our problem

1-p is the probability of fail: 0.25 in our problem

Then, substitute:

$P(X=x)=C(15,x)\cdot(0.75)^x\cdot(0.25)^{(15-x)}$

That is the frequency function.

If you want to give it as a table you must find P(X=1), P(X=2), P(X=3), . . . up to P(X=15) using that function. That is not part of the question.

6 0

2 years ago

Caleb is a football enthusiast. He watches all the big games, collects football memorabilia, and knows everything there is to kn

kotykmax [81]

Answer:

enduring;

situational

Explanation:

The concept of enduring participation implies the existence of a long-term relationship, regardless of the current situation.

Situational participation depends on the moment and how attractive the event is in a specific period of time.

6 0

3 years ago

A. On April 1, the company hired an attorney for a flat monthly fee of $500. Payment for April legal services was made by the co

uranmaximum [27]

Answer and Explanation:

The journal entries are shown below:

1 Legal fees expense $500

To Legal fees payable $500

(Being accrued expense is recorded)

2 Legal fees payable $500

To Cash $500

(Being amount paid is recorded)

3 Interest expense $2,007

To Interest payable $2,007

(Being accrued interest is recorded)

4 Interest payable $2,007

Interest expense $4,013

To Cash $6,020

(Being interest paid is recorded)

5 Salaries and wages expense $11,200

To Salaries and wages payable $11,200

(Being accrued expense is recorded)

6 Salaries and wages payable $11,200

Salaries and wages expense $2,800

To Cash $14,000

(Being salaries and wages paid is recorded)

8 0

3 years ago

A company manufactures and sells x smartphones per week. The weekly price-demand and cost equations are p=500−0.5x and C(x)=20,0

suter [353]

Answer:

revenue = x amount of phones x price

revenue = x(500 - 0.5x)

revenue = 500x - 0.5x²

we find revenue' (derivative):

revenue' = 500 - x

x = 500

the company should sell 500 smartphones to maximize revenue, the selling price = 500 - (0.5 x 500) = $250 per smartphone. Maximum weekly revenue = $250 x 500 = $125,000

profit = revenue - cost

profit = 500x - 0.5x² - 20,000 - 135x

profit = -0.5x² + 365x - 20,000

we must find profit' (derivative):

profit' = -x + 365

x = 365

In order to maximize profits, you have to sell 365 smartphones per week. Maximum weekly profit = -0.5(365²) + 365(365) - 20,000 = -66,612.50 + 133,225 - 20,000 = $46,612.50.

The smartphone's price = 500 - (0.5 x 365) = $317.50

4 0

2 years ago