Let the groups be A, B, C and D. Then
n(A ∩ B ∩ C ∩ D) = 1
n(A ∩ B ∩ C ∩ D') = 10 - 1 = 9
n(A ∩ B ∩ C' ∩ D) = 10 - 1 = 9
n(A ∩ B' ∩ C ∩ D) = 10 - 1 = 9
n(A' ∩ B ∩ C ∩ D) = 10 - 1 = 9
n(A ∩ B ∩ C' ∩ D') = 100 - 9 - 9 - 1 = 81
n(A ∩ B' ∩ C ∩ D') = 100 - 9 - 9 - 1 = 81
n(A' ∩ B ∩ C ∩ D') = 100 - 9 - 9 - 1 = 81
n(A' ∩ B ∩ C' ∩ D) = 100 - 9 - 9 - 1 = 81
n(A' ∩ B' ∩ C ∩ D) = 100 - 9 - 9 - 1 = 81
n(A ∩ B' ∩ C' ∩ D) = 100 - 9 - 9 - 1 = 81
n(A ∩ B' ∩ C' ∩ D') = 1000 - 81 - 81 - 81 - 9 - 9 - 9 - 1 = 729
n(A' ∩ B ∩ C' ∩ D') = 1000 - 81 - 81 - 81 - 9 - 9 - 9 - 1 = 729
n(A' ∩ B' ∩ C ∩ D') = 1000 - 81 - 81 - 81 - 9 - 9 - 9 - 1 = 729
n(A' ∩ B' ∩ C' ∩ D) = 1000 - 81 - 81 - 81 - 9 - 9 - 9 - 1 = 729
Thus, all together there are 4(729) + 6(81) + 4(9) + 1 = 2,916 + 486 + 36 + 1 = 3,439 members in the groups.
Therefore, there are all together 3,439 members in the groups.
Answer:
¡hola Emma! ¡Soy Hazel, encantado de conocerte!
Step-by-step explanation:
Answer:
d. 5
Step-by-step explanation:
I just learned how to do it just for this so please lmk if it's right or not :)
Answer:
The standard deviation for the sample mean distribution=0.603
Step-by-step explanation:
We are given that
Mean,
Standard deviation,
n=44
We have to find the standard deviation for the sample mean distribution using Central Limit Theorem for Means.
Standard deviation for the sample mean distribution
Using the formula
Hence, the standard deviation for the sample mean distribution=0.603
Answer:B
Step-by-step explanation:
