Based on this question, one thing that we would seriously consider would be the fact of first, doing

first. By doing this, it would then give us our answer as 16. By us understanding this point of view, we would then consider that this would then be your answer. That would then include that there would then be 16 pairs of the "enantiomeric pairs", and that would then be the possible estimate.
<span>a.2
b.4
c.8
d.16</span>
You have the correct answer being (-5,3). Simply add the parentheses around the point
Answer:
0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Mean of 0.6 times a day
7 day week, so 
What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

In which




So

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.
ANSWER
A) 9.4 * 10^-8, 9.25 * 10^-6, 2.5 * 10^3, 7* 10^3
EXPLANATION
The numbers are given in standard form.
The first criteria we will use to order them is the exponents.
The bigger the exponents the bigger the number.
The second criteria is that, if the exponents of any two numbers are the same, then we use the numbers multiplying the powers of 10 to order.

The correct choice is A.
The x variable, that is being multiplied.