Kayla's expected value for a 2- point shot is 1 point.
What is expected value?
The expected value, also known as expectation, expectancy, arithmetical expectation, mean, average, first or moment, is a generalization of the weighted average in the field of probability theory. Informally, the estimation is the arithmetic mean of a significant number of values of a random variable that were independently chosen. A weighted average of all potential outcomes constitutes the expected value of the a random variable with such a finite number of outcomes. When there is a continuum of potential outcomes, integration defines the expectation. The expectation is provided by Lebesgue integration in the measure theory-based axiomatic foundation for probability.
Sol- The expected value is calcuated using the formula:
EV= x.P(x)
where x is the weight/value of the event, and P(x) is the probability of the event occurring.
For a 2-point shot, we have:
X=2
P(x) = 50/100 = 0.5
There for the expected value will be -
EV = 2×0.5 = 1
Thus,
The expected value will be 1 point.
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Answer:
the awnser would be a(g(x)=3x2
F* you B*!!!!!! Your so S*! That's the easiest thing in the world!!
The answer to this question is
B, (-infinity, -28]. We can get this answer by first multiplying each side of the inequality by 7. That would get rid of the fraction. When one does that, the result is d + 28
0. That means that d
-28. In interval notation, which is the notation the problem is asking us, that would be
(-infinity, -28], since d is all values less than -28 this includes infinity, but it also includes -28, so there is a ] around it. That means that the answer to this question is
B, (-infinity, -28].
Answer:
f(1)=34 ; f(n)= f(n-1) +6, for n >2
Step-by-step explanation:
arithmetic sequence
