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True [87]
2 years ago
15

Please help this is due tomorrow. 1/2 of 20?

Mathematics
2 answers:
SashulF [63]2 years ago
8 0
Divide 20 by 2 and you get the answer = 10

\frac{20}{2}  = 10
Mars2501 [29]2 years ago
6 0

1/2 of 20 is 20/2 which is 10.

All you really need to do here is to divide 20 by 2 to get 10.


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Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $3,034 was collected on the sale of 1,226 ticket
Mila [183]
452 adult tickets were sold.
774 student tickets were sold.
6 0
3 years ago
If a/\2+1/a/\2=11, then a+1/a=?
Nadya [2.5K]

Answer:

\pm \sqrt{13}

Step-by-step explanation:

\bigg(a +  \frac{1}{a}  \bigg)^{2} =  {a}^{2}   +  \frac{1}{ {a}^{2} }  + 2 \times a \times  \frac{1}{a}  \\  \\ \bigg(a +  \frac{1}{a}  \bigg)^{2} = 11 + 2 \\  \\   \bigg(a +  \frac{1}{a}  \bigg)^{2} = 13 \\  \\  \bigg(a +  \frac{1}{a}  \bigg) =  \pm \sqrt{13}

5 0
3 years ago
April has 95 baseball cards.She wants to organize them on pages that hold 18 cards each. She has 5 pages. Does April have enough
gulaghasi [49]
Hello, 

we must multiply the number of pages by the number of cards that hold each page to find the total number of card that she can organize, so:

5*18=90

She can organize 90 cards, but she has 95, then SHE DOESN'T HAVE ENOUGHT PAGES TO ORGANIZE THEM.
8 0
2 years ago
Read 2 more answers
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:

Expected value = 40/26 = 1.54 approximately

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
Does anybody want to talk?
aleksandrvk [35]

Answer:

yes please

finals are killing me and i'm homeschooled so this is just hell

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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