Answer:
Mike spends $160,000 on A and $75,000 on B
Explanation:
Lets call the rental properties as A and B.
A yields 10% anb B yields 12%.
Total investment=$235,000
Expected income=$25,000
Let the amount he invests in A be x.
Then amount spent in B=$(235,000-x)
Income from A=x × 0.1 ----------------(1)
Income from B=(235,000-x)× 0.12 ---------------(2)
Sum of incomes =$25,000
Taking sum of both equations,
x × 0.1 +(235,000-x)× 0.12 =25,000
235,000×0.12 -0.02×x =25,000
28,200 -0.02×x =25,000
0.02×x =3,200
x=
x =$160,000
Answer:
taxes and no money management
Explanation:
some comes out of taxes and you do not know what to do with so much money
Answer:
Incremental cost= $61,875
Explanation:
Giving the following information:
Gelb Company currently manufactures 49,500 units per year of a key component for its manufacturing process. Variable costs are $5.15 per unit, fixed costs related to making this component are $75,000 per year, and allocated fixed costs are $70,500 per year. The allocated fixed costs are unavoidable whether the company makes or buys this component. The company is considering buying this component from a supplier for $3.90 per unit
We need to determine whether it is more convenient to produce the component or outsource it. We will only consider the relevant costs, therefore the fixed costs will not be taken into account.
Make in house:
Cost= 49,500*5.15= $254,925
Buy:
Cost= 49,500*3.90= $193,050
Incremental cost= 254,925 - 193,050= $61,875
Complete question:
Lovely lawns, inc., intends to use sales of lawn fertilizer to predict lawn mower sales. the store manager estimates a probable six-week lag between fertilizer sales and mower sales. the pertinent data are:
Period Fertilizer Sales (tons) Number of Mowers Sold (six-week lag)
1 1.4 9
2 1 7
3 1.5 10
4 1.8 12
5 2.1 13
6 1.5 7
7 1.3 5
8 1.2 5
9 1.6 8
10 1.3 7
11 1.6 11
12 1.3 9
13 1.4 10
14 1.8 12
a. Graph these data to see whether a linear equation might describe the relationship between fertilizer and mowers.
b. Obtain a linear regression line for the data
c. Predict expected lawn mower sales for Period 15, given fertilizer sales six weeks earlier of 2.3 tons.
Solution:
a. Find the attachment for graph
b.
Fertilizer Mowers
x y
1.4 9 1.96 81 12.6
1 7 1 49 7
1.5 10 2.25 100 15
1.8 12 3.24 144 21.6
2.1 13 4.4 116 27.3
1.5 7 2.25 49 10.5
1.35 5 1.69 25 6.5
1.2 5 1.44 25 6
1.6 8 2.56 64 12.8
1.3 7 1.69 49 9.1
1.6 11 2.56 121 17.6
1.3 9 1.69 81 11.7
1.4 10 1.96 100 14
1.8 12 3.24 144 21.6
Σ =20.8 125 31.94 1201 193.3
x = 1.486
y = 8.929
EXY- nXY 193.3- (149.486 -1.486)
Calculated using the formula b =9 = 31 .94 - (14 *1. 486 *1 A86 ) =7.31405
X2 - 7.31405
Calculated using Excel a =Y - bX = 8.929 - 7.31405 *1.486 = -1.938017
Y = a + bx = -1.93802 + 7.31405x
c. Using the formula
Y = a + bx = -1.58678 + 7.033058x
= 14.8843
Using Excel's Forecast
Forecast - 14.58926
D) the availability of land, labor and capital
I think