The
chance to land on the number three is only a ¼ or 0.25 chance. Knowing that we
can also understand the law of large numbers. The law of large numbers states
that as the number of spins gets very large, eventually the experimental
probability will reach the theoretical probability. Eight spins of the spinner
is not a large number, so the probabilities will be different. In this case
resulting in zero times landing on three. Therefore, Megan should not be
concerned that the spinner is unfair.
Since, the probability of success during a single event of a geometric experiment is 0.34.
We have to find the probability of success on the 6th event.
Since it is a geometric experiment. So, when a discrete random variable 'X' is said to have a geometric distribution then it has a probability density function (p.d.f.) of the form:
P= 
, where q = 1 - p
So, now
P =
where 'p' is the probability of success and 'q' is the probability of failure and x is the number of events.
Since the probability of success (p)is 0.34
Therefore, probability of failure(q)= 1 - p
= 1 - 0.34
= 0.66
and x = 6
So, P =
= 
= 
= 0.0425
So, the nearest tenth of a percent of probability of success on the 6th event =
4.257 %
Rounding to the nearest tenth, we get
= 4.3%
So, Option A is the correct answer.
Step-by-step explanation:
Let y1 and y2 be (e^x)/2, and (xe^x)/2 respectively.
The Wronskian of them functions be
W = (y1y2' - y1'y2)
y1 = (e^x)/2 = y1'
y2 = (xe^x)/2
y2' = (1/2)(x + 1)e^x
W = (1/4)(x + 1)e^(2x) - (1/4)xe^(2x)
= (1/4)(x + 1 - x)e^(2x)
W = (1/4)e^(2x)
Since the Wronskian ≠ 0, we conclude that functions are linearly independent, and hence, form a set of fundamental solutions.
Answer:
15 27 44
Step-by-step explanation the two smaller numbers combined has to be bigger than the other number.
