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Nikolay [14]
3 years ago
5

Using Exhibit 2–6, Typical After-Tax Budget Allocations for Different Life Situations, calculate the budget allocations for Jami

e Lee using her net monthly salary (or after-tax salary) amount. Is she within the recommended parameters for a student?

Business
1 answer:
NikAS [45]3 years ago
6 0

Answer:

Refer below for the explanation.

Explanation:

Jamie lee is inside all the suggested parameters set for an understudy life in spending costs of housing, transportation, savings, and entertainment. Her sparing is marginally over the suggestion of 10%, with her ascertaining proportion at 11%.

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Luebke Incorporated has provided the following data for the month of November. The balance in the Finished Goods inventory accou
vodka [1.7K]

Answer: $‭238,800‬

Explanation:

Adjusted Cost of Goods for November = Beginning Finished good inventory + Cost of goods manufactured  - Ending Finished goods inventory - Overapplied Overheads

Overapplied Overhead = Overhead applied - Actual Overhead

= 60,400 - 56,800

= $3,600

Adjusted Cost of Goods for November = 58,000 + 215,000 - 30,600 - 3,600

= $‭238,800‬

8 0
3 years ago
A company's product is very complex, and there is value in having employees develop deep expertise in each aspect of the product
Drupady [299]

The organizational structure that would allow for deep, specialized knowledge to be fostered is a<u> functional structure. </u>

<h3>What is a functional organization structure?</h3>

This refers to a practice of many companies where people who have different expertise are put into different departments that complement their skills.

This ensures that the employees will develop deep expertise in their field because they will focus on it by being in a department that is based on their expertise.

Find out more on functional organizational structures at brainly.com/question/14700402.

4 0
3 years ago
Suppose now that market demand for skiing increases to Qᴅ = 9000 − 60p because of environmental regulations neither Pepall Ridge
jeyben [28]

Answer:

they both produce the same thing

Explanation:

check the picture attached below for the full explanation.

8 0
3 years ago
Consider once again the project with the cash flows described below. Determine whether this is a simple or non- simple investmen
barxatty [35]

Answer:

The correct option is (D)

Explanation:

Using the accumulated cash flow sign test,

- This investment is a non-simple investment.

This is because the net cash flow changes sign (to positive and back to negative) more than once, during the study period.

- There are at most 3 i* values. That is values for Internal Rate of Return. This is owing to the multiple change in sign during the period.

The correct answer is option D - Non simple investment with at most 3 values for internal rate of return (i*)

5 0
3 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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