(X-3)*log(1/2)=(x-5)*log(1/4)
X-3=(log(1÷4)÷log(1÷2))*(x-5)
X-3=2(x-5)
X-3=2x-10
X-2x=3-10
-x=-7
X=7
Answer:
x_>-4
Step-by-step explanation:
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).
That should be answer no.a
18.33-12.25=6.08
16.55-9.48=7.07
7.07 > 6.08
so the answer should be no.a
15% of 1950 = 1950 * 0.15 = 292.5
So, sum would be: 1950 + 292.5 = 2242.5
In short, Your Answer would be $2242.5
Hope this helps!
