For a probability distribution to be represented, it is needed that P(X = 0) + P(X = 1) = 0.44. Hence one possible example is:
<h3>What is needed for a discrete random variable to represent a probability distribution?</h3>
The sum of all the probabilities must be of 1, hence:
P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) = 1.
Then, considering the table:
P(X = 0) + P(X = 1) + 0.15 + 0.17 + 0.24 = 1
P(X = 0) + P(X = 1) + 0.56 = 1
P(X = 0) + P(X = 1) = 0.44.
Hence one possible example is:
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Answer:
-32
Step-by-step explanation:
The absolute value of a number is its distance from 0. Since distance is always positive the absolute value is the same number but positive. So the absolute value of -32 and 32 would both be 32 because they have the same distance from 0.
Answer:
92
Step-by-step explanation:
Given
2n + t² ← substitute n = - 4 and t = - 10 into the expression
= 2(- 4) + (- 10)² = - 8 + 100 = 92
Answer:
Step-by-step explanation:
x+y=10
x=10-y
x=10-5
x=5
3y= -3x+30
3(10-y)= -3(10-y)+30
30-3y= -30+3y+30
30+30-30=6y
30=6y
y=5
Answer:
assuming 60 and 63 are the side lengths
87 in is the hypotenuse
Step-by-step explanation:
60^2 + 63^2 = x^2
