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tia_tia [17]
3 years ago
5

. Find the sum of the geometric sequence. (1 point) 1, one divided by four, one divided by sixteen, one divided by sixty four, o

ne divided by two hundred and fifty six
Mathematics
2 answers:
il63 [147K]3 years ago
8 0

Answer:

0.332

Step-by-step explanation:

given series

1/4, 1/16,1/64.1/256

this is geometric series

where common ratio   r is given by

nth term/ (n-1)th term

let the second term is nth term and first term is (n-1)th term

r = 1/16 / (1/4) = 1/4

___________________________________________

sum of series is given by

a (1-r^n)/1-r

where a is first term

n is the number of terms

r is the common ration

___________________________________________

in the given series

1/4, 1/16,1/64.1/256

a = 1/4

r = 1/4

n = 4

thus ,

sum = 1/4(1-(1/4)^4)/ (1-1/4)

sum = 1/4(1-(1/256)/(4-1)/4

sum = 1/4((256-1)/256 / 3/4

1/4 in numerator and denominator gets cancelled

sum =( 255/256*3) = 255/768 = 0.332

Thus, sum of series is 0.332.

vlabodo [156]3 years ago
6 0

Answer:

341/256

Step-by-step explanation:

I took the test and got the answer right

You just give all the fractions a common denominator of 256 and then change and add up the numerators and you get 341

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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
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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.

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