D is the answer, just add al the sides (multiply each by two and add them)
9514 1404 393
Answer:
2 2/3
Step-by-step explanation:
Let the number be represented by x. Then the problem statement is telling you ...
4 5/9 = 2 1/3 + (5/6)x
You can isolate the x term by subtracting 2 1/3 from both sides of the equation. (The properties of equality require that the same operation be performed on both sides of the equation.)
(4 5/9) - (2 3/9) = (2 3/9) -(2 3/9) +(5/6)x
2 2/9 = (5/6)x
The definition of the multiplicative inverse tells you that the coefficient of x can be made to be 1 if the term is multiplied by (6/5). We must do that to both sides of the equation.
(6/5)(2 2/9) = (6/5)(5/6)x
(6/5)(20/9) = x . . . . simplify
120/45 = x . . . . . . do the multiplication
8/3 = x = 2 2/3 . . . . . . . reduce the fraction
The number is 2 2/3.
_____
<em>Additional comment</em>
The number we're looking for is multiplied by 5/6 and the result has 2 1/3 added to it. To find the number, we "undo" these operations, in reverse order. We undo the addition by subtracting the amount that was added. We undo the multiplication by multiplying by the inverse of that factor.
This is the sort of logic that you would use to fill in the blank for ...
7 = 1 + 3×___
You probably recognize that this breaks down into two problems:
7 = 1 + ___ . . . . . (6 goes in the blank)
and
6 = 3×___
You may notice that for all of these problems, a good knowledge of addition and multiplication facts is useful.
-(4 1/5)
5(4)+1 = 21
Thus the answer would be -21/5
The coefficient C-term is 1 in problem. The coefficient is the number before the variable of said equation. Example, the coefficient of X for thus problem 4x-4y is 4 the number BEFORE the variable, never variable itself it won’t ever be 4x
Answer:
Step-by-step explanation:
Given that:
X(t) = be the number of customers that have arrived up to time t.
... = the successive arrival times of the customers.
(a)
Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;
Now 
(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;
Now; 
(c) Determine the conditional probability density function for W2, given that X(t)=5.
So ; the conditional probability density function of 
given that X(t)=5 is:
