Answer:
200-20=180
Step-by-step explanation:
Answer:
Where's the image?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
Since the warranty on a machine specifies that it will be replaced at failure or age 4 and the distribution for X is defined between 0 and 5 then if we define the random variable Y ="the age of the machine at the time of replacement" we know that the values for Y needs to be between 0 and 4 or between 4 and and we can define the following density function:
for other case
Now we can apply the definition of expected value and we have this:
And for the second moment we have:
And the variance would be given by:
Answer:
4,5,6
Step-by-step explanation:
(x)+(x+1)+(x+2)= 15
3x + 3 = 15
3x = 12
x = 4
Answer:
48
x
2
−
184
x
+
192
Step-by-step explanation: