For a discrete probability distribution, you add up x*P(x) over all possible values of x, where P(x) is the probability that the random variable X takes the value x. For a continuous distribution you need to integrate x*P(x) with respect to x.
"x2 = 100" is an equation which, if solved for x, will give you two possible value: 10 and -10.
Given a value for the variable x, you find (calculate) the corresponding value of y. These (x, y) pairs are part of the table. You cannot complete the table because there are infinitely many possible values of x.
the value of x is 3
zero. The absolute value of a number is just the positive version of that number, so the absolute value of x is x, and x minus x is zero.
There is no possible value of x that will satisfy the given equation. There is no possible value of x that will satisfy the given equation. There is no possible value of x that will satisfy the given equation. There is no possible value of x that will satisfy the given equation.
If x is the unknown or variable in an equation it can have many possible maximum or minimum values
Positive X or Negative X
Given a situation, what are the possible values of X is what it is asking.
It is not possible to be certain. If x >= 0 then abs(x) = x if x < 0 then abs(x) = -x
There is no such value since rational and irrational numbers are infinitely dense. If there was a positive number, x, that laid claim to having the smallest possible absolute value, it would immediately be challenged by x/2. Then that, in turn, would be challenged by x/4, and then x/8 and so on, and on, and on.
The only possible value of p is 3.
Both increment the value of the variable by one. The difference is the value of the increments expression itself. With preincrement value is taken after incrementing, and with postincrement value is taken before incrementing. Example: Let x have value 5. y = ++x; Both y and x are assigned value 6. Again let x have value 5. y = x++; y is assigned value 5. x is assigned value 6.
Without an equality sign the given terms can't be classed as an equation and so therefore finding the value of x is not possible.
For a discrete probability distribution, you add up x*P(x) over all possible values of x, where P(x) is the probability that the random variable X takes the value x. For a continuous distribution you need to integrate x*P(x) with respect to x.
One possible inequality that has x = 0.8 as a solution is x ≤ 0.8. This means that any value of x that is less than or equal to 0.8 will satisfy the inequality.
-1 < x < 64, but if x is less than 0 then x can be any value.