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solniwko [45]
3 years ago
5

Suppose you bought a house for $3,250,000 to make it a nursing home in the future. But you have not committed to the project and

will decide in nine years whether to go forward with it or sell off the house. If real estate values increase annually at 1.5%, how much can you expect to sell the house for in nine years if you choose not to proceed with the nursing home project?
Business
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

$3,716,050

Explanation:

FV = PV × (1 + i)∧n

Present Value (PV) 3250000  

Interest Rate (i) 0.015  

Number of years (n) 9

   (1 + 0.015) ∧ 9

        3,250,000 x 1.1434

       =$3,716,050

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Automobile repair costs continue to rise with the average cost now at $367 per repair.† Assume that the cost for an automobile r
vovikov84 [41]

Answer:

a)  0.1728

b)  0.09183

c) 0.7354

d) $ 222.25

Explanation:

Given

mean = \mu = $367

Standard deviation = \sigma =$88

Cost of automobile repair is normally distributed.

a) We have to find P( x > 450 )

P( x > 450 ) = 1 - P( x <= 450 )

Using excel function,   P( x <= x ) = NORMDIST (x,  \mu, \sigma, 1 )

P( x > 450 )   = 1 - NORMDIST( 450 , 367, 88, 1 )

= 1 - 0.8272 = 0.1728

P( x > 450 ) = 0.1728

b)  P( x < 250 ) = NORMDIST( 250 , 367, 88, 1 ) = 0.09183

P( x < 250 ) = 0.09183

c) P( 250 < x < 450 ) = P( x <450 ) - P( x < 250 )

P( x <450 ) = NORMDIST( 450 , 367, 88, 1 ) = 0.8272

P( x < 250 ) = NORMDIST( 250 , 367, 88, 1 ) = 0.09183

P( 250 < x < 450 ) = 0.8272 - 0.09183 = 0.7354

P( 250 < x < 450 ) = 0.7354

d) We have P( X < a ) = 0.05

We have to find a.

Using Excel, = NORMINV ( Probability, \mu, \sigma )

a = NORMINV ( 0.05 , 367, 88 ) = 222.2529

Cost = $ 222.25

8 0
3 years ago
Find the future values of these ordinary annuities. Compounding occurs once a year. Round your answers to the nearest cent. $200
PIT_PIT [208]

Answer:

Normal:

$ 3,509.7470

$    563.7093

$ 2,000.00

Due:    

 $3,930.9167

 $   597.5319

 $ 2,000.00

Explanation:

We solve using the formula for common annuity and annuity-due on each case:

C \times \frac{(1+r)^{time} }{rate} = FV\\

C \times \frac{(1+r)^{time} }{rate}(1+rate) = FV\\ (annuity-due)

<u>First:</u>

C 200.00

time 10

rate 0.12

200 \times \frac{11+0.12)^{10} }{0.12} = FV\\

200 \times \frac{11+0.12)^{10} }{0.12}(1+0.12) = FV\\

Normal:  $3,509.7470

Due:       $3,930.9167

<u>Second:</u>

100 \times \frac{(1+0.06)^{5} }{0.06} = FV\\

100 \times \frac{(1+0.06)^{5} }{0.06} (1+0.06)= FV\\

$563.7093

$597.5319

<u>Third:</u>

No interest so no time value of money the future value is the same as the sum of the receipts regardless of time or being paid at the beginning or ending.

1,000  + 1,000 = 2,000

4 0
2 years ago
The per-unit standards for direct labor are 2 direct labor hours at $15 per hour. If in producing 2900 units, the actual direct
pashok25 [27]

Answer:

$5400 Favorable

Explanation:

Standard 2 hour at $15 per hour

Standard hours 2 hour per unit * 2900 units = 5800 hours

Total Standard cost = 5800 hours * $15 per hour =  $87,000

Actual hours = 5100

Actual cost = $81600 / 5100 hours = $16 per hour

Variance = Standard - Actual

Labor hour Variance Favorable = 700 hours (5800 hours - 5100 hours)

Total Labor variance = $5400 ($87,000 - $81,600)

4 0
3 years ago
Suppose in the short run a firm’s production function is given by Q = L 1 2 K 1 2 and that K is fixed at K = 10. If the price of
Furkat [3]

The firm’s marginal cost of production when the firm is producing 50 units of output is 33.33

Solution:

The production function is Q = \sqrt{L * K}

The initial value is 10 units. The production value is 50 units The manufacturing cycle needs work as stated below.

Q = \sqrt{L * K}

Q = \sqrt{L * 10}

L = (\frac{Q}{3.162} )^{2}

The wage rate is $15 . The following is the expense of the manufacturing process.

TC = P_{L} * L + P_{K} * K

TC = ( 15 * (\frac{Q}{3.162} )^{2} ) + [ P_{k * 10}]

The marginal production cost is really the increase in manufacturing costs as output increases by 1 point.

As listed below, the marginal cost:

TC = ( 15 * (\frac{Q}{3.162} )^{2} ) + [ P_{k * 10}]

MC = \frac{TC}{Q} = \frac{2Q}{3}

MC = \frac{2*50}{3} = 33.33

6 0
2 years ago
What is a modification problem? What are the three possible types of modification problems?
Oxana [17]

Answers and explanations:

1) A modification problem takes places when creating a database two different type of information is entered in the same chart row generating inaccuracy. The only form to solve this issue is creating a new row so each piece of information will be stored in one row particularly.

2) There are three (3) types of modification problems: the deletion problem (<em>the single row containing information from different themes can be deleted losing data</em>), the update problem (<em>new information entered could lead to more inconsistency</em>), and the insertion problem (<em>similar to deletion, a new row can be inserted instead of the row causing problem but information will be missing</em>).

4 0
2 years ago
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