economic factors such as supply and demand can play an


 

economic factors such as supply and demand can play an important role in your investment process. If there is a huge supply of houses available in a neighborhood you like, housing prices in that neighborhood may go down due to a lack of current demands. Therefore, you as the buyer have more power and may be able to make an offer that is lower than the initial asking price. Alternatively, in a neighborhood that is very much in demand and low in available housing, the housing prices may be higher, and you may need to offer to pay more to compete with other offers for a home there.

Decisions in investments can end in a variety of ways. Applying your problem solving skill and critical thinking strategies can help you successfully analyze opportunities to bring the best benefits for your efforts. Adjusting for the time value of money lets you calculate if the rewards are worth the wait so that you can arrive at an informed decision before you commit your money.

In this week’s discussion: 

Identify an investment decision that you or someone else has made. (This should be for goods or services for which one or more payments were made in order to receive increased benefits from those goods or services at a later time.)

  • How do you think the concept of supply and demand impacted the investment decision?
  • Now that you understand the time value of money, how would you use your problem solving skill (especially critical thinking strategies) to rethink the decision in the future?

Share This Post

Email
WhatsApp
Facebook
Twitter
LinkedIn
Pinterest
Reddit

Order a Similar Paper and get 15% Discount on your First Order

Related Questions

Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′ − 4y′ + 3y = x ; use y1 = e3x vi. y′′ + 5y′ + 6y = e2xcos(x) vii. y′′ + y = sec(x) tan(x)

Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′