Answer:
The multiple choices are:
$5,589.04
$7,452.05
$4,890.41
$5,876.71
$6,410.96
Amount invested in K is $6,410.96
Explanation:
L+K=12,000
from the return perspective
0.0975=K/12000*0.0805+L/12000*0.117
K=12000-L
Substitute for K in the second equation
0.0975=(12000-L)/12000*0.0805+L/12000*0.117
0.0975=(966-0.0805L)/12000+0.117L/12000
0.0975=(966-0.0805L+0.117L)/12000
12000*0.0975=966+0.0365
L
1170
-966=0.0365L
204=0.0365L
L=204/0.0365
L=$ 5,589.04
K=$12,000-$ 5,589.04
K=$6,410.96
Answer: Starbucks Coffee is a 'normal good', while Beanlightened coffee is an 'inferior good'.
Andrew's demand for Starbucks coffee changed as a result of an increase in his 'income'
Explanation:
A normal good is a good that sees it's demand rise as income or wages rise. Essentially if you're making more money, you buy more of such goods. Andrew is now making more money so he buys more of Starbucks coffee.
An inferior good on the other hand is one that sees it's demand drop as wages or income rises. You usually buy less of it the more money you make. Take no brand cornflakes for instance, as one makes more money they tend to buy less of it and more of branded cornflakes. Beanlightened coffee is therefore an inferior good.
Income is compensation you get for providing a service. In this instance Andrew receives $75000 a year for being a programmer.
Answer:
$6
Explanation:
depreciation rate per hour using the units-of-production method = (cost of asset - residual value) / estimated hours of operation
($80,000 - $5,000) / 12,500 = $6
Answer: Product departmentalization
Explanation: In simple words, product departmentalization refers to a process in which an organisation puts all the activities related to a project under a single manager. All the activities relating to that product will be performed in that separate department.
In the given case. the organisation is dividing all their products in separate departments. Hence we can conclude that they most likely follows product departmentalization.
Correct being that he is in high school!<span />
