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FrozenT [24]
3 years ago
14

14. If a business process, and information relating to that process, cannot be patented, copyrighted, or trademarked, there is n

othing a business can do to protect it. a. True b. False 15. Which of the following is NOT a category of trade secret
Business
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

False

Explanation:

The reason is that the business can protect such type of information by keeping it patented. So the safety measure would be to keep it secret and if the information is leaked the company can sue the employees who breached the confidentiality and also by having an agreement with the organization to keep it secrect and divide the markets. So the company can protect if it is determined to protect such information.

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Jessica is a U.S. Army Reservist and in 2020 traveled 130 miles each way to serve duty at a local military installation. She was
Zepler [3.9K]

Answer:

AGI Deduction = 580.8 US dollars.

Explanation:

Solution:

Data Given:

Miles traveled = 130 miles

Reporting Period = 4 times in 2020

Cost from home to base = $1.75 toll each way.

Jessica's for AGI deduction for these costs is:

First, we need to find the travel amount:

Travel amount = Miles x Reporting Period x 2 x Mileage rate for deduction

Here, mileage rate for deduction = 0.545 per mile. So,

Travel Amount = 130 miles x 4 times x 2 x 0.545

Travel Amount = 566.8 US dollars.

Now, we need to find the Jessica's Toll Expenses.

Toll Expenses = 1.75 x 2 x 4 times

Toll Expenses = 14 US dollars.

Finally, we can find the required AGI Deduction:

AGI Deduction = Travel Amount + Toll Expenses.

AGI Deduction = 566.8 US dollars + 14 US dollars

Hence, Jessica's AGI deductions are:

AGI Deduction = 580.8 US dollars.

3 0
3 years ago
Youngstown Construction plans to discontinue its roofing segment. Last year, this segment generated a contribution margin of $65
ipn [44]

Answer:

B. a decrease of $30,000

Explanation:

The computation of company’s overall profit is shown below:-

To continue = Contribution margin - Fixed cost

= $65,000 - $70,000

Loss = $5,000

To Discontinue =  Unavoidable fixed cost ÷ 2

= $70,000 ÷ 2

= $35,000

So, Net Loss = To continue (Loss) - To Discontinue

= $5,000 - $35,000

= $30,000

Therefore there is a decrease of $30,000

7 0
3 years ago
"A borrower obtained a $10,000 term loan with 6 1/2% interest paid yearly. A $1,000 principal reduction was to be paid with each
aleksley [76]

Answer:

$1585

Explanation:

Interest for the first year = 6.5% of principal due at the beginning of the year

= 6.5% of $10,000

= $ 650

Principal repayment at the end of the year = $1000

Principal due at the beginning of the second year = $10,000 - $1000= $9000

Interest payable at the end of the second year = 6.5% of principal outstanding at the beginning of the second year = 6.5% of 9000

                                                                            = $ 585

Principal repayment at the end of the second year = $1000

Hence total payment at the end of the second year = $1000 + $585= $1585

6 0
3 years ago
Hopefully, the pattern is clear! Call the amount owed after the payment at the start of Monthn An. Based off ofthis pattern, wri
Llana [10]
Cucufate gudhfhfjjrjjryeywjejfbfjgj ejgj xbxhfdggHugh HD jfjfjjf ur tu f
8 0
4 years 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
3 years ago
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