Answer:
2/7
then 2/14
Step-by-step explanation:
Let P(H)=p be the probability of one head. In many scenarios, this probability is assumed to be p=12 for an unbiased coin. In this instance, P(H)=3P(T) so that p=3(1−p)⟹4p=3 or p=34.
You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.
The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together: P(H)P(H)P(H)=p3 or in your case, (34)3=2764.
Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is 3(p2)(1−p). The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: HHT,HTH,THH. In your case (where the coin is 3 times more likely to have heads): 3(34)2(14)=2764.
Adding those events together you get p3+3(p2)(1−p)=5464. Note that the 3 coefficient
Y=14x+9 the values x=1,2,3
Answer:
Step-by-step explanation:
I clicked see solution
T=5 years
12% of 72 is:
=12×72/100
=8.64
So the 12% of 72 is 8.64 is your answer.
We can try reduction order and look for a solution 
. Then
Substituting these into the ODE gives
which leaves us with an ODE linear in 
:
This ODE is separable; divide both sides by the coefficient of 
and separate the variables to get
Integrate both sides; on the right, substitute 
so that 
.
Now solve for 
,
then for 
,
Solve for 
by integrating both sides.
Substitute again and solve for 
:
then for ,
So the second solution would be
already accounts for the second term of the solution above, so we end up with
as the second independent solution.
