I am not sure about this but here goes nothing: d=2.5t+2.2 is very familiar to y=mx+b y=d m=2.5 x=t b=2.2 to find a new quasion we substitute d and t into place wich equals 1=2.5(0)+b simplification and you get b=1 so your answer is d=2.5t+1
Answer:
.0625 or A
Step-by-step explanation:
You have to work out the mean. (The simple average of those numbers) Then for each number subtract the Mean and square the result. Then work out the average of those squared differences.
Answer:
0.34285714285 i think
Step-by-step explanation:
Answer:
V(t) = 25000 * (0.815)^t
The depreciation from year 3 to year 4 was $2503.71
Step-by-step explanation:
We can model V(t) as an exponencial function:
V(t) = Vo * (1+r)^t
Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.
We have that Vo = 25000, r = -18.5% = -0.185, so:
V(t) = 25000 * (1-0.185)^t
V(t) = 25000 * (0.815)^t
In year 3, we have:
V(3) = 25000 * (0.815)^3 = 13533.58
In year 4, we have:
V(4) = 25000 * (0.815)^4 = 11029.87
The depreciation from year 3 to year 4 was:
V(3) - V(4) = 13533.58 - 11029.87 = $2503.71
I would go with the second statement is true because if all of the other ones mentioned in the problem had something that would not work either the one before or after would make the statement false excepted for statement two
