Answer:
a. 2 years
b. 1 year
c. 12 times
Explanation:
Interest period is the duration of the deposit. It is the length of time the money would remain in deposit. This is 2 years according to the question
Compounding period = number of times interest would be paid. In the question, this is a year. So interest would be paid every year
The compounding frequency - it is the number of times the deposit would be compounded. It is 12 months
The future value of the deposit can be determined using this formula :
FV = P (1 + r/m)^nm
FV = Future value
P = Present value
R = interest rate
N = number of years
m = number of compounding
Answer:
The correct answer is A. The Mars Climate Orbiter crashed on the surface of Mars because one program output thrust in terms of foot-pounds, and another program expected thrust to be expressed in terms of newtons.
Explanation:
The Mars Climate Orbiter was NASA’s unsuccessful mission to study the Martian climate, part of the Mars Surveyor 98 program. The MCO was created as a satellite-translator for the Mars Polar Lander lander, and after the latter ceased to function, it was supposed to study the Martian climate.
The Mars Climate Orbiter was destroyed when a navigation error caused the probe to be improperly elevated as it entered orbit. The vehicle was destroyed by the friction and stresses in Mars' atmosphere. An investigation showed that some data for the rocket system were calculated in English units (pound-force-second) while the navigation team expected SI units (Newton-second).
Answer:
Hart's note should be reported at $10,000 and Maxx's note should be reported at $7,820
Explanation:
Since Hart's note is a current note (due within one year) it should be reported at future value = $10,000
Marxx's note must be reported at present value:
present value = future value x discount factor = {$10,000 [1 + (3% x 5)]} x 0.68
present value = $11,500 x 0.68 = $7,820
*we use simple interest to calculate the future value of Marxx's debt since Jet Co. doesn't charge compound interest
Answer:
Explanation:
Assumed Data
Budgeted Sales 1000000
units sold 10000
Unit price 100
Cost Per unit 60
Before After Cahnge Due to
impelemtation implementation implementation
Sales 1000000 1125000* 125000
Cost -600000 -750000 -150000
Profit 400000 375000 -25000
Advertise Cost 0 -30000 -30000
400000 345000 -55000
* Sales price 100
Reduction 10%
After Reduction Sp 90
Current unit sales 10000
Increase 25%
After increase 12500
Cost Per unit will remain the same because only sales price will be decreased to boost the sale
New sales 12500*90 1125000
Cost 12500*60 750000
