Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on three of its faces, and the number 3 painted on one of its faces. Assume that each face is equally likely to come up. If the die were loaded so that the face with the 3 on it were twice as likely to come up as each of the other five faces, would this change the value of P(odd number)

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

And the probability of each is:

P(Odd Number) is then calculated as:

Take LCM

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

The probability of each is:

Take LCM

Comparing P(Odd Number) before and after

--- Before

--- After

We can conclude that the change to the face 3 affects the value of P(Odd Number)

60

step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204