a. Let be a random variable representing the weight of a ball bearing selected at random. We're told that
, so
where . This probability is approximately
b. Let be a random variable representing the weight of the
-th ball that is selected, and let
be the mean of these 4 weights,
The sum of normally distributed random variables is a random variable that also follows a normal distribution,
so that
Then
c. Same as (b).
Answer:
See explanation
Step-by-step explanation:
Ms. Johnson space has 48 carpet squares.
Mr. Wei space has 36 carpet squares.
Factorize these two numbers:
All common factors are 1, 2, 3, 4, 6, 12
<u>All possibilities:</u>
1. If Ms. Johnson space width and Mr. Wei space width is 1 carpet, then Ms. Johnson space length is 48 carpets and Mr. Wei space length is 36 carpets.
2. If Ms. Johnson space width and Mr. Wei space width is 2 carpets, then Ms. Johnson space length is 24 carpets and Mr. Wei space length is 18 carpets.
3. If Ms. Johnson space width and Mr. Wei space width is 3 carpets, then Ms. Johnson space length is 16 carpets and Mr. Wei space length is 12 carpets.
4. If Ms. Johnson space width and Mr. Wei space width is 4 carpets, then Ms. Johnson space length is 12 carpets and Mr. Wei space length is 9 carpets.
5. If Ms. Johnson space width and Mr. Wei space width is 6 carpets, then Ms. Johnson space length is 8 carpets and Mr. Wei space length is 6 carpets.
6. If Ms. Johnson space width and Mr. Wei space width is 12 carpets, then Ms. Johnson space length is 4 carpets and Mr. Wei space length is 3 carpets.