You haven't provided the series, therefore, I can only help with the concept.
<u><em>For an infinite geometric series, we have two possibilities for the common ratio (r):</em></u>for r > 1, the terms in the series will keep increasing infinitely and the only possible logic summation of the series would be infinity
for r < 1, the terms will decrease, therefore, we can formulate a rule to get the sum of the infinite series
<u><em>In an infinite series with r < 1, the summation can be found using the following rule:</em></u>sum =

where:
a₁ is the first term in the series
r is the common ratio
<u>Example:</u>
For the series:
2 , 1, 0.5 , 0.25 , ....
we have:
a₁ = 2
r = 0.5
Therefre:
sum =

Hope this helps :)
Answer:
0.25
Step-by-step explanation:
72% of courses have final exams and 46% of courses require research papers which means probability of 0.72 for courses that have final exams and 0.46 for courses that require research papers.
31% of courses have a research paper and a final exam, which means probability of 0.31 for both courses with exams and research papers, using Venn diagram approach, find picture attached to the solution.
P(R or E) = P(R) + P(E) - P(R and E), which gives:
P(R or E) = 0.15 + 0.41 - 0.31
P(R or E) = 0.25.
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
Answer:

f(x) = 4 when x is 8
Step-by-step explanation:
Domain is the set of x values that make the function defined. Allowed x values for the function (mapping).
The Range is the set of y values that make the function defined. Allowed y values for the function (mapping).
- Whenever we need to find f(a), suppose, then we look for "a" in the domain and see its corresponding value mapping in the range.
- Whenever we will be given a value for f(x) = a, suppose, and we have to find "x", we look at the value a in the range and find corresponding x value in the domain.
Firstly, we need f(4), so we look for "4" in domain and see which number it corresponds to in range.
That is 
Thus,

Next,
We want "x" value that gives us a "y" value of 4. We look for "4" in the range and see which value it corresponds to. That is "8". So,
f(8) = 4