Answer:
The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%
Step-by-step explanation:
Given
Before 1995
(Br) Brown = 30%
(Y) Yellow = 20% =0.2
(R) Red = 20%
(G) Green =10% =0.1
(O) Orange = 10%
(T) Tan = 10%
After 1995
(Br) Brown = 13%
(Y) Yellow = 14% =0.14
(R) Red = 13%
(G) Green = 20%
= 0.2
(O) Orange = 16%
(Bl) Blue = 24%
Since there are two bags, let A be the bag from 1994, and B be the bag from 1996
Then let AY imply we drew a yellow M&M from the 1994 bag
AG implies we drew a green M&M from the 1994 bag
BY implies imply we drew a yellow M&M from the 1996 bag
BG implies we drew a green M&M from the 1996 bag
P(AY) =0.2
P (BY) = 0.14
P(AG) =0.1
P(BG) =0.2
Since the draws from the 1994 and 1996 bag are independent,
therefore

The draws can happen in either of the 2 ways in (1) and (2) above
therefore total probability E is given as
E =
For the yellow one to be from 1994, it implies that the event to be chosen is

Since the total probability is given as E=0.054
then 
Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E
Answer:
Step-by-step explanation:
Hypotheses:

(Left tailed test)
p = sample proportion
n = sample size = 200
Sample proportion = p = 73/200 = 0.365
Std error = 0.03404
p difference = 0.365-0.39 = 0.025
t = 0.025/se = 0.7344
df = 199
p value = 0.2317
Since p >0.05 accept null hypothesis
No,the veterinarian does not have a right to be skeptical because there is no statistical evidence to support his claim.
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There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy
•
divides 
•
divides 
I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

Since
divides
, and

where
is degree-2, and we can write it as

Now, we have

so if
divides
, then
is degree-2, so
is degree-4, and we can write

where
is degree-1.
Expanding the left side gives

and dividing by
leaves no remainder. If we actually compute the quotient, we wind up with

If the remainder is supposed to be zero, then

Adding these equations together and grouping terms, we get

Then
, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

So, we end up with
