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e-lub [12.9K]
3 years ago
9

What are three consecutive integers that add up to 4,590

Mathematics
1 answer:
miskamm [114]3 years ago
7 0
1529+1530+1531=4590

All you have to do is divide 4590 by three and you get 1530. After that add 1 and you get 1531. Then go back to 1530 and subtract 1 and you get 1529 and there are your three consecutive integers that equal 2590.(1529, 1530,1531)
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Which angle pairs are supplementary? Check all that apply.
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You divide a number and get a remainder of 4. what 1 - digit number can the divisor of the problem?
ahrayia [7]

Answer:

<em>Any of the numbers 5,6,7,8,9 could be the divisor of the problem.</em>

Step-by-step explanation:

<u>Division</u>

If I divide a number and get a remainder of 4, then the divisor must be greater than 4. A divisor of 4, for example, would lead to a remainder of (maximum) 3.

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Your friend asks if you would like to play a game of chance that uses a deck of cards and costs $1 to play. They say that if you
gtnhenbr [62]

Answer:

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The player expects to win on average about $1.54 per game.

The positive expected value means it's a good idea to play the game.

============================================================

Further Explanation:

Let's label the three scenarios like so

  • scenario A: selecting a black card
  • scenario B: selecting a red card that is less than 5
  • scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

  • 1/2 = 13/26
  • 2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

13/26 + 36/26 - 9/26

(13+36-9)/26

40/26

1.538 approximately

This value rounds to 1.54

The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.

Therefore, this game is tilted in favor of the player and it's a good decision to play the game.

If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).

Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

7 0
2 years ago
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