Answer: Your answer is r^12t^12
Step-by-step explanation: You use the distribution property.
Answer and Explanation:
Solution: The operation of concatenation for a set of string on p. and the set is
AB = {XY | X ∈ A and y ∈ B}.
We need to satisfy all these following properties to find out the standard set is closed under concatenation.
1- Union of two standard sets also belongs to the classic collection. For example, A and B are regular. AUB also belongs to a regular group.
2- Compliment of two standards set A and B are A’ and B’ also belonging to the standard set.
3- Intersection of two standards set A and B is A∩B is also a regular set member.
4- The difference between two regular sets is also standard. For example, the difference between A and B is A-B is also a standard set.
The closure of the regular set is also standard, and the concatenation of traditional sets is regular.
For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Answer:
0.000025
Step-by-step explanation:
PEMDAS (Exponets first.)
10^5 = 100,000
10^10 = 10,000,000,000
rewrite
2.5 x 100,000 divided by 1.0 x 10,000,000,000
PEMDAS (multiplication next.)
2.5 x (10^5) = 250,000
1.0 x (10^10) = 10,000,000,000
rewrite.
250,000 divided by 10,000,000,000 = 0.000025
0.000025
