Answer:
ok but helppp meeeee nothinggggg
Answer:
mkana
Step-by-step explanation:
Since the basis is from year 1 to year 2, calculate first for the difference of their percentages. That would be:
Difference = year 2 - year 1
Difference = 2.32% - 1.1% = 1.22%
We apply this same value of percentage increase from year 2 to year. Thus, the percentage for year 3 is:
% Year 3 = % Year 2 + percentage increase
% Year 3 = 2.32% + 1.22%
% Year 3 = 3.54%
Well. If you get 29% of 5 cars that is 145%. This was a little confusing but I am defiantly sure it is correct. I like to use simpler numbers to see if I am doing the work right. So I said if he has a 50 % likely hood to find a car that was expired and had 1 car. It would be 50 percent. Now if he had 2nt got it the first time it would be a 100 % chance to find the car expired . Hope I didn’t co fuse you more
Theory:
The standard form of set-builder notation is <span>
{ x | “x satisfies a condition” } </span>
This set-builder notation can be read as “the set
of all x such that x (satisfies the condition)”.
For example, { x | x > 0 } is
equivalent to “the set of all x such that x is greater than 0”.
Solution:
In the problem, there are 2 conditions that must
be satisfied:
<span>1st: x must be a real number</span>
In the notation, this is written as “x ε R”.
Where ε means that x is “a member of” and R means “Real number”
<span>2nd: x is greater than or equal to 1</span>
This is written as “x ≥ 1”
Answer:
Combining the 2 conditions into the set-builder
notation:
<span>
X =
{ x | x ε R and x ≥ 1 } </span>
