The possible transfer prices that could be used on transfers between the Windshield and Assembly divisions is $200 to $450.
The first piece of glass you see on most cars is the windshield. Also known as European car windshield. Windshields play an important role in supporting the vehicle structure and protecting the driver and passengers. The windshield protects occupants from wind, dust, insects, rocks and other flying objects and provides an aerodynamically shaped front panel. By applying UV coating, you can block harmful ultraviolet rays.
However, most car windows are made of laminated safety glass, so this is usually unnecessary. Most of the UV-B is absorbed by the glass itself and the remaining UV-B is absorbed by the PVB tie layer along with most of the UV-A.
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Answer:
The average operating cost is $0.46 per mile
In deciding whether to to her use her own car or rent a car the costs are analysed below:
Variable operating cost is a relevant cost
Depreciation is not relevant as it is already cost and also it is sunk cost
insurance is not relevant as well
automobile tax and license is not relevant as it would be paid regardless of the option chosen
Explanation:
The average cost comprises of the variable operating cost per mile as well as the fixed operating cost per mile
variable operating cost per mile is $0.06
fixed cost operating cost=fixed costs/total miles driven=($3,350+$1,700+$900+$450)/16000=$6400
/16000=$0.40
average cost per mile=$0.06+$0.40=$0.46
Based on the scenario above, the economic concept which Frakie is faced with is OPPORTUNITY COST. Opportunity cost refers to a benefit or value that a person could have received but which he gave up in order to take another course of action. Thus, an opportunity cost represents an alternative given up when a decision is made.
Answer:
(a) 
(b) 
(c) X=4.975 percent
Explanation:
(a) Find the z-value that corresponds to 5.40 percent
.
Hence the net interest margin of 5.40 percent is 2.5 standard deviation above the mean.
The area to the left of 2.5 from the standard normal distribution table is 0.9938.The probability that a randomly selected U.S. bank will have a net interest margin that exceeds 5.40 percent is 1-0.9938=0.0062
(b) The z-value that corresponds to 4.40 percent is 
The net interest margin of 4.40 percent is 0.5 standard deviation above the mean.
Using the normal distribution table, the area under the curve to the left of 0.5 is 0.6915
Therefore the probability that a randomly selected U.S. bank will have a net interest margin less than 4.40 percent is 0.6915
(c) The z-value that corresponds to 95% which is 1.65
We substitute the 1.65 into the formula and solve for X.
A bank that wants its net interest margin to be less than the net interest margins of 95 percent of all U.S. banks should set its net interest margin to 4.975 percent.
