Looking at the set, we are given 18 elements. 17 is prime; it has only two factors: 1 and 17, since 1•17=17. So, the question is really asking what is the probability the numbers 1 or 17 is chosen. As mentioned earlier, 17 is prime, so there are two possible choices: 1 and 17.
P (probability) = possible outcomes / total outcomes
It is important to note that these events are “or” events, meaning that the probability can only be determined by choosing a 1 or a 17; you can’t randomly chose a 1 and 17 at the same time. So, the formula is:
P(A or B) = P(A) + P(B)
All this is saying is that given two possible outcomes, the probability occurs independent of each event; they don’t occur at the same time.
P(1 or 17) = P(1)/18 + P(1)/18
P(1 or 17) = 2/18
Since 17 is prime, it’s two and only factors are 1 and 17. The probability of randomly choosing a 1 or 17 is 2/18, meaning that there are 2 elements in the set out of a possible 18 elements that can be randomly chosen.
2/18 simplifies to 1/9
So, your answer is 1/9
The right answer for the question that is being asked and shown above is that: "B. X^2 = 5.85 p = 0.88." The answer that would be the appropriate type of test to investigate our hypothesis is that <span>"B. X^2 = 5.85 p = 0.88." This is the correct answer.</span>
So the function is
the intial value is where x=0
where x=0, the value of f(0)=3
grouth is 1/3 but it is decay, so it might be correct, or not because it is decay
it is exponential decay
true it is a stretch of the funciton f(x)=(1/3)^x
when x=3, then f(3)=1/9
the true things are
The function shows exponential decay.
The function is a stretch of the function f(x) = (1/3)x .
I believe the answer would be B. But im not sure if it was read right because its sideways
