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Sphinxa [80]
2 years ago
11

A company is starting a fish farm in a man-made lake with a carrying capacity of 7500. The CEO wants to have a population of 500

0 fish after 3 years. If the yearly population growth rate is 2.0, about how many fingerlings should they add to the lake to start their population?
Business
1 answer:
Cloud [144]2 years ago
5 0

Answer:

4,712 fingerlings

Explanation:

The amount that a sum will accumulate to at the end of a particular number of period if it grows at a certain rate per annum is given as

F= A× (1+g)^(n)

F- Sum at the end of the period, A- sum at the beginning of the period, n- number of period, g -growth rate

5,000= A× (1.02)^(3)

a= 5000/(1.02^3)

a= 4711.61

The  population at the start of fingerlings = 4,712

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The maintenance expenses on a rental house you own average $200 a month. The house cost $219,000 when you purchased it four year
Serhud [2]

Answer:

value we place on this house when analyzing the option of using it as a professional office is $225000

Explanation:

Given data

house cost 4 year ago  = $219,000

house valued = $239,000

real estate fees = $14000

property taxes = $4,000

to find out

What value should you place on this house

solution

we know if we sell house we should pay real estate fee

so we get need money to place is present cost - real estate fees

so cost will be

cost = house valued  - real estate fees

cost = 239000 - 14000

cost = 225,000

so value we place on this house when analyzing the option of using it as a professional office is $225000

0 0
2 years ago
An appliance dealer must decide how many (if any) new microwave ovens to order for next month. The ovens cost $220 and sell for
Vlada [557]

Answer:

Explanation:

Order 0: we have unsold items for which the return is -25

return is -25*(.4*1+.2*2+.1*3) = -25*1.1 = $-27.50

Order 1: we have to sell at a discount if no orders, otherwise sell 1, and unsold items if demand 2 or 3

return is .3*(1/2*300-220) + (1-.3)*(300-220) + -.25*(.2*1+.1*2) = .3*-70+.7*80+-25*(.4) =

-21 + 56 - 10 = $25

Order 2: we have to sell at a discount if 0 or 1 orders, sell 1 or 2, and unsold items if demand 3

return is (.3*2+.4*1)*(1/2*300-220)+(.4*1+(.2+.1)*2)*(300-220)+-25*.1 =1*-70+1*80-25*.1 =

-70 + 80 - 2.5 = $7.50

Order 3:

return is (.3*3+.4*2+.2*1)*(1/2*300-220)+(.4*1+.2*2+.1*3)*(300-220) = 1.9*-70 + 1.1*80 =

-133 + 88 = -$45

Order 1, with a return of $25, as this is the highest return.

b) If we had a perfect information, we would never pay a penalty for underordering or suffer a discounted return from over-ordering

(.4*1+.2*2+.1*3)*(300-220) = 1.1*80 = $88

Then, the value of perfect information is $88 - $25 = $63

c) P(D=0|F) = P(F|D=0)*P(D=0)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.1*.3/(.1*.3+.2*.4+.3*.2+.9*.1)=.03/.26 = 3/26

P(D=1|F) = P(F|D=1)*P(D=1)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.2.4/(.1*.3+.2*.4+.3*.2+.9*.1)=.08/.26 = 4/13

P(D=2|F) = P(F|D=2)*P(D=2)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.3*.2/(.1*.3+.2*.4+.3*.2+.9*.1)=.06/.26 = 3/13

P(D=3|F) = P(F|D=3)*P(D=3)/(P(F|D=0)*P(D=0)+P(F|D=1)*P(D=1)+P(F|D=2)*P(D=2)+P(F|D=3)*P(D=3))=

.9*.1/(.1*.3+.2*.4+.3*.2+.9*.1)=.09/.26 = 9/26

P(D=0|U) = P(U|D=0)*P(0)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.8*.3/(.8*.3+.3*.4+.1*.2+.1*.1)=.24/.39 = 8/13

P(D=1|U) = P(U|D=1)*P(1)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.3*.4/(.8*.3+.3*.4+.1*.2+.1*.1)=.12/.39 = 4/13

P(D=2|U) = P(U|D=`)*P(`)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.1*.2/(.8*.3+.3*.4+.1*.2+.1*.1)=.02/.39 = 2/39

P(D=3|U) = P(U|D=3)*P(3)/(P(U|D=0)*P(D=0)+P(U|D=1)*P(D=1)+P(U|D=2)*P(D=2)+P(U|D=3)*P(D=3))=

.1*.1/(.8*.3+.3*.4+.1*.2+.1*.1)=.01/.39 = 1/39

P(N|D=0 = 1-.1-.8 = .1

P(N|D=1) = 1 - .2 - .3 = .5

P(N|D=2) = 1 - .3 - .1 = .6

P(N|D=3) = 1 - .9 - .1 = 0

P(D=0|N) = P(N|D=0)*P(D=0)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))=.1*.3/(.1*.3+.5*.4+.6*.2+.0*.1)= .03/.35 = 3/35

P(D=1|N) = P(N|D=1)*P(D=0)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))= .5*.4/(.1*.3+.5*.4+.6*.2+.0*.1)= .20/.35 = 4/7

P(D=2|N) = P(N|D=2)*P(D=2)/(P(N|D=0)*P(D=0)+P(N|D=1)*P(D=1)+P(N|D=2)*P(D=2)+P(N|D=3)*P(D=3))= .6*.2/(.1*.3+.5*.4+.6*.2+.0*.1)= .12/.35 = 12/35

P(D=3|N) = 0

If the result of the survey is an F, we have

P(D=0|F) = 3/26

P(D=1|F) = 4/13

P(D=2|F) = 3/13

P(D=3|F) = 9/26

If the order is 0, the return is -25*(1*4/13+2*3/13+3*9/26) = -25*47/26 = -1175/26 = -$45.19

If the order is 1, the return is 3/26*-70+(1-3/26)*80+-25*(1*3/13+2*9/26) = 515/13 = $39.62

If the order is 2, the return is (3/26*2+4/13)*-70+(1*4/13+2*(3/13+9/26))*80 + -25*9/26 =

1835/26 = $70.58

If the order is 3, the return is (3/26*3+4/13*2+3/13)*-70+(1*4/13+2*3/13+3*9/26)*80 =

795/13 = $61.15

We should order 2.

P(D=0|U) = 8/13

P(D=1|U) = 4/13

P(D=2|U) = 2/39

P(D=3|U) = 1/39

If we order 0, the return is (4/13*1+2/39*2+1/39*3)*-25 = -475/39 = -$12.18

If the order is 1, the return is 8/13*-70+(1-8/13)*80+-25*(1*2/39+2*1/39) =-580/39= -14.87

If the order is 2, the return is (8/13*2+4/13)*-70+(1*4/13+2*(2/39+1/39))*80 + -25*1/39 =

-2785/39= -$71.41

If the order is 3, the return is (8/13*3+4/13*2+2/39*1)*-70+(1*4/13+2*2/39+3*1/39)*80 =

-1780/13 = -$136.92

Order 0

P(D=0|N) = 3/35

P(D=1|N) = 4/7

P(D=2|N) = 12/35

P(D=3|N) = 0

If we order 0, the return is (4/7*1+12/35*2)*-25 = -220/7 = -$31.43

If the order is 1, the return is 3/35*-70+(1-3/35)*80+-25*(1*12/35) = 410/7 = $58.57

If the order is 2, the return is (3/35*2+4/7)*-70+(1*4/7+2*12/35)*80 = 340/7 = $48.57

We don't order 3, as the probability of 3 is 0

we order 1

We order 2 if there is an F, 0 if there is an N, and 1 if there is a U.

d) P(F) = .26

P(N) = .39

P(U) = .35

Then, the expected return is .26*1835/26 +-475/39*.39 + 410/7*.35 = $34.10

Since we make $25 if we just take 1, we should pay up to $34.10-$25 = $9.10 for the survey.

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