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Jobisdone [24]
2 years ago
8

Annie's team has just finished a major project and the team has, after a long time, got some free time on hand. However, for the

last couple of days, Annie has noticed that few of the team members are immersed in online social networking sites during office hours. 7.
What is the least effective responsive?
(A) Annie arranges for a team meeting, thanking them for all the hard work put in and reminding them that, in spite of having some free time on hand, they are to use it productively and not indulge in non-productive activities like visiting online social networking sites during office hours. She also shows them additional online training courses they could attend instead on the company's intranet, creates a training roster, and circulates to the team.
(B) Annie sends out a mail to the team thanking them for all the hard work put in and reminding them that, in spite of having some free time on hand, they are to use it productively and not indulge in online social networking sites during office hours. She also attaches the company policy regarding the same in her email.
(C) Annie talks to the offenders one-on-one and reminds them that, in spite of having some free time on hand, they are to use it productively and not indulge in online social networking sites during office hours. She also offers to show them additional online training courses they could attend instead on the company's intranet.
Business
1 answer:
Natasha_Volkova [10]2 years ago
5 0

Answer:

B.

Explanation:

B is giving a bit of a impersonal approach, and comes across as just pushing company rules, especially by attaching a company policy.  Her response should be more precise with options.  Annie does not provide that in response B.

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Richard participated in a study conducted by an advertising agency. During his interview, he was asked to gauge the services pro
Rasek [7]

Answer:

The correct answer is letter "A": product-specific preplanning inputs.

Explanation:

Product-specific preplanning input is a series of efforts carried out by advertisements agencies to collect information about<em> industry competitors, work processes, and consumers patterns and preferences </em>on determined products that will allow them to create a strategy to merchandise a new good or service in the market.

Product-specific preplanning input makes use of <em>focus group interviews and demographic and psychographic segmentation </em>as feeds to create marketing strategies.

6 0
3 years ago
During December, Rainey Equipment made a $658,000 credit sale. The state sales tax rate is 6% and the local sales tax rate is 1.
crimeas [40]

Answer:

Debit: Accounts Receivable 707,350

Credit: Sales Revenue 658,000

Credit: Sales taxes payable ([6% + 1.5%] × $658,000) = $49,350

Explanation:

3 0
3 years ago
The Smelting Department of Kiner Company has the following production data for November. Production: Beginning work in process 3
Paul [167]

Answer and Explanation:

The computation of the equivalent units of production for both material and the conversion cost is shown below:

Particulars                                           Materials         Conversion costs

Unit transferred out                                9,700                            9,700

Add:

Ending work in process                         8,300                            3,818

                                                     (8300 × 100%)               (8,300 × 46%)

Total equivalent unit                               18,000                         13,518

6 0
3 years ago
Libby Company purchased equipment by paying $6,700 cash on the purchase date and agreed to pay $6,700 every six months during th
vladimir1956 [14]

Answer:

The answer is $53,732.

Explanation:

The value of the equipment reported on Libby Company's balance sheet is equal to:

Cash payment at purchase + Present value of 8 equal semiannual payment, $6,700 each discounted at 3% ( because semiannual payment is made for 4 years so we have 2 x4 = 8 payments; and annual borrowing rate is 6% so we have discount rate = 6% /2 = 3%).

with:

Cash payment at purchase = $6,700;

Present value of 8 equal semiannual payment, $6,700 each discounted at 3% = (6,700/3%) x ( 1 - 1.03^(-8) ) = $47,032 ( that is, apply the formula to find present value of annuity).

we have:

The value of the equipment reported on Libby Company's balance sheet = 6,700 + 47,032 = $53,732.

5 0
2 years ago
The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almana
storchak [24]

Answer:

a) P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

b) P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

c) P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the scores for critical reading of a population, and for this case we know the distribution for X is given by:

X \sim N(502,100)  

Where \mu=502 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(502-10 < \bar X < 502+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

Part b

Let X the random variable that represent the scores for Math of a population, and for this case we know the distribution for X is given by:

X \sim N(515,100)  

Where \mu=515 and \sigma=100

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)

And for this case we want this probability:

P(515-10 < \bar X < 515+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

Part c

Let X the random variable that represent the scores for Writing of a population, and for this case we know the distribution for X is given by:

X \sim N(494,100)  

Where \mu=494 and \sigma=100

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)

And for this case we want this probability:

P(494-10 < \bar X < 494+10)

And for this case we can use the z score given by:

z= \frac{\bar X -\mu}{\sigma_{\bar x}}

And if we use this formula we got:

P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

3 0
3 years ago
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