Answer:
This took me a long time to figure out , but I'm pretty sure number 1 is
D) Silk and Wool.
And number 2 is C) Northern Nigeria
Explanation: Hope that's right, sorry if not because for number 1 most of them could be the answer, and for number 2, three of them could be the answer.
The expected return on this portfolio will be given by:
E[P]=Rf+(E[Rm]-Rf)β
Where:
Rf=Risk Free interest rate
Rm=Return on the market portfolio
β= Market Beta
The return on our portfolio will be:
E[p]=0.043+(0.128-0.043)0.013
=0.043+0.085*0.013
=0.044105
=4.4105%
Answer:
$6250
$5000
$5250
Explanation:
Straight line depreciation expense = (Cost of asset - Salvage value) / useful life
($212,000 - $12,000) / 8 = $25,000
The machine was used for only 3 months in the fiscal year. Thus, the depreciation expense = $25,000 x (3/12) = $6250
Activity method based on output = (output produced that year / total output of the machine) x (Cost of asset - Salvage value)
(1000 / 40,000) x ($212,000 - $12,000) = $5000
Activity method based on hours worked = (hours worked that year / total hours of the machine) x (Cost of asset - Salvage value)
($212,000 - $12,000) x (525 / 20,0000) = $5250
Answer:
The Silverside Company
Project 1's Payback Period
= Initial Investment/Annual cash flows
= $400,000 / $90,000
= 4.44 years.
Explanation:
Project 1:
Initial Investment = $400,000
Useful life = 5 years
Annual cash inflows for useful life = $90,000
The Silverside Company's payback period calculates the time or number of years that it would take the company to recover from its initial investment in Project 1. This is the simple payback period calculation. There is also the discounted payback period calculation. This method discounts the annual cash inflows to their present values before the calculation is carried out. This second method gives a present value perspective on the issue.
Answer: V=7.43m/s
d =2.82m
Explanation:
a) For the first part, the initial velocity immediately after ejection, by using momentum conservation
before ejection, the momentum of the squid/water system is zero
there are no external forces acting on the system at the moment of ejection, so we can find the speed of the squid by noting
momentum before ejection = momentum after ejection
0 = M1U + M2V
0=-0.26 kg x 20 m/s + 0.7kg x V
where the speed of the water is taken as the negative sign, and V is the speed of the squid right after ejection, solving for V we get
V=7.43m/s
B. we use the equation vf^²=v0^²+2ad
where vf=final velocity = 0 since velocity is zero at motion's apex
v0=initial velocity = 7.43m/s
a = acceleration = -9.8m/s/s
d=height (to be found)
Therefore,
0=7.43^²+2(-9.8)d
Mathematically, it becomes
d=7.43^²/2(9.8)= 2.82m
d = 2.82m
