Agame is played using one die. if the die is rolled and shows 5 5, the player wins $ 10 $10. if the die shows any number other than 5 5, the player wins nothing. if there is a charge of $ 2 $2 to play the game, what is the game's expected value?

The expected value is -$0.33

Step-by-step explanation: The expected value is calculated as:

E = ∑pₙxₙ

where xₙ is a given event (in this case winning 8$ or losing 2$) and pₙ is the probability for the event.

So in this you only win with one result in the dice, so the probability of winning is 1/6, and the probability of losing must be 5/6.

Then the expected value is:

E = 1/6*$8 - 5/6*$2 = $-0.33

Where I used the profit of $8 instead of $10, because you must pay $2 for play, and you win $10, the profit is ($10 - $2) = $8

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation:

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake