A.) Since there is not interest, there is no time value of money to be recovered. That means that the $100 today, would still be $100 a few years from now. The balance function would then be:
B(p) = $100 - 20p
b.) Since the half of 100 is 50, equate B(p) to 50
B(p) = 50 = 100 - 20p
p = 2.5
Therefore, after 3 payments, Jenna have paid back more than half of the loan.
c.1) The same procedure is done as that of part a. However, the base amount is $120 instead of $100, and the $20 is replaced with $15. The new equation becomes:
B(p) = 120 - 15p
c.2) To determine the time, let the balance zero out.
0 = 120 - 15p
Solving for p,
p = 8
Thus, Jenna could pay back the loan after 8 months.
I believe the answer is... 4 3/4
Answer:
No because its -10 per year so after 10 years the car value will go below zero which makes it useless
Step-by-step explanation:
Answer:
0.9177
Step-by-step explanation:
let us first represent the two failure modes with respect to time as follows
R₁(t) for external conditions
R₂(t) for wear out condition ( Wiebull )
Now,
where t = time in years = 1,
n = failure rate constant = 0.07
Also,
where t = time in years = 1
where Q = characteristic life in years = 10
and B = the shape parameter = 1.8
Substituting values into equation 1
Substituting values into equation 2
let the <em>system reliability </em>for a design life of one year be Rs(t)
hence,
Rs(t) = R1(t) * R2(t)
t = 1
![Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713](https://tex.z-dn.net/?f=Rs%281%29%20%3D%20%5Be%5E%7B-0.07%7D%20%5D%20%2A%20%5Be%5E%7B-0.0158%7D%20%5D%20%3D%200.917713)
Rs(1) = 0.9177 (approx to four decimal places)
Yes, the table is correct
There would be 60 squares as the ratio of triangles to squares is 3:4
