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).
iAnswer: i belive it is 12
Step-by-step explanation: theGCF of 36 and 24 is 12 the GCF of 36 and 48 is 12 and the GCF of 48 and 24 is 12 and 24
PRT TWO: there would be 2 pencils in each bag 3 candy barsin each and 4 erasers in each because 24 divided by 12 = 2 36 divided by 12 = 3 and 48 dividedby 12 = 4
<span>A+B)^2 is the largest. It is A^2+2AB+B^2, which is clearly greater than the last two options. To compare (A+B)^2 and 2(A+B), we remove one A+B so that we're just comparing A+B and 2. As A+B must be at least 3 (as both must be positive integers, and one must be greater than the other, leading to a minimum value of A=2, B=1), A+B is greater than 2, and as a result, (A+B)^2 is always the largest.</span>
Answer:
its A
Step-by-step explanation:
i literally just took the test