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nika2105 [10]
2 years ago
6

When preparing for a business trip to China, Kaylee Putbrese determined she needed to bring $5,200. How much must she borrow for

a simple discount note at 6% for 45 days?
Business
2 answers:
Musya8 [376]2 years ago
5 0

Answer:

Explanation:

The question is confusing with regards to what exactly is being asked. it is not clear whether we should calculate the amount she must borrow now if she wants to pay $4500 in 45 days given a 6% note or How much would she pay in 45 days if she borrows $5200 today. I have provided solution for all scenarios.

Value = 4500

r = 0.06/365 = 0.00016484

n = 45 days

Let X be the the amount she must borrow now

Value = X(1 + rn)

4500 = X(1 + 0.00016484 x 45)

4500 = X(1.0074178)

X = 4500/1.0074178 = 4466.8379729

X = 4466.84.

if she wants to pay the exact $4500 she must borrow $4466.84 now

or if she wants to take $4500 she must invest $ 446.84 on a 6% note 45 days.

Future Value of 4500 in 45 days

Future Value = 4500(1 + 0.00016484 x 45) = 45333.3801

Future Value = 45333.38

she would need to pay $45333.38 in 45 days if she takes $4500 now for her trip

Sergeeva-Olga [200]2 years ago
3 0

Answer:

Borrowed amount = $5417

Explanation:

Discount note = 4%

This means that Kaylee has 100 - 4 = 96% of the borrowed amount at hand

Cash at hand = $5,200

Let the borrowed amount = X

Cash at hand = 96% of X

5200 = (96/100) * X

X = (5200 * 100)/96

X = $5417

Borrowed amount = $5417

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The manager at East Coast Manufacturing organizes costs to prepare the Costs of Quality report. The manger compiled the followin
Angelina_Jolie [31]

Answer:

The total cost of quality the manager should use to report the costs in the internal failure cost​ category is $297,000

Explanation:

The computation of the internal failure cost is shown below:

=  Rework cost + Cost of rejected units

= $257,000 + $40,000

= $297,000

The cost of internal failure includes both the cost of rework and the rejected units.

The other information which is given in the question is not relevant. Hence, ignored it as it would not be considered and thus not taken in the computation part.

4 0
2 years ago
At the beginning of the month, the Painting Department of Skye Manufacturing had 20,000 units in inventory, 70% complete as to m
velikii [3]

Answer:

Cost per EUP Materials: $2,00

Cost per EUP Conversion: $4,50

Explanation:

                                              Materials                        Conversion

Beginning WIP Cost              $22,400                             $6,250

Cost added in the period   <u>$229,600                         $540,500</u>

Total Cost of the Units        $252,000                         $546,750

Equivalent Units of Production (EUP): Completed Units + Ending WIP Units

EUP Materials = 120,000 + 15,000 x 40% = 126,000

EUP Conversion = 120,000 + 15,000 x 10% = 121,500

Cost per Equivalent Unit: Cost of Units / EUP  

Cost per EUP Materials: $252,000 / 126,000 = $2,00

Cost per EUP Conversion: $546,750 / 121,500 = $4,50

5 0
1 year ago
Yvette is a customer of Apexon Bank, which is a member of the FDIC. She currently has a checking account that has $17,371 and a
GaryK [48]
Yvette has a checking account with $17,371 and a savings account with $240,000. Her combined money in Apexon Bank is $257,371. 

To know how much of Yvette's money is protected you must note that:
FDIC insures: checking, savings, money market deposits and certificates of deposit. FDIC protects against $250,000 combined. 

Since Yvette has $257,371 the FDIC protects against $250,000 of that amount leaving $7,371 unprotected. 
4 0
2 years ago
A delivery service is buying 600 tires for its fleet of vehicles. One supplier offers to supply the tires for $ 80 per​ tire, pa
nignag [31]

Answer:

$4,372.71

Explanation:

Here for reaching the difference in PV between the first and the second offer first we need to follow some steps which is shown below:-

Step 1

Total payment due = Per tire × Bought tires

= $80 × 600

= $48,000

Step 2

Present value factor of 8.4% for 1 year = 1 ÷ (1 + Rate of interest)^Number of years

= 1 ÷ (1 + 8.4%)^1

= 1 ÷ (1 + 0.084)^1

= 1 ÷ 1.084

= 0.92251

Step 3

First offer

Present value = Total payment due × Present value factor of 8.4% for 1 year

= $48,000 × 0.92251

= $44,280.48

Step 4

Second offer

One year payment = Bought tires × Per tire

= 600 × $45

= $27,000

Step 5

Present value = One year payment × Present value factor of 8.4% for 1 year

= 27,000 × 0.92251

= $24,907.77

Step 6

Total present value = Present value of second offer + Tires cost

= $24,907.77 + $15,000

= $39,907.77

Here we can see that first offer is higher than second offer

So,

The difference between the first and the second offer = First offer - Second offer

= $44,280.48 - $39,907.77

= $4,372.71

7 0
2 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
2 years ago
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