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3 years ago

Can anyone figure this out?

Mathematics

1 answer:

Verizon [17]3 years ago

5 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

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2 years ago

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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.

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2 years ago

Ella has 0.5 lbs of sugar. How much water should she add to make the following concentrations? Tell Ella how much syrup she will

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Answer:

1) 2 lbs

2) 0.17 lbs

3) 9.5 lbs

Step-by-step explanation:

Amount of sugar Ella has = 0.5 lbs

Let the amount of water she adds to make the solution = x lbs

So total amount of the mixture would be = (x + 0.5) lbs

Case 1: 20% syrup

20% syrup means the sugar percentage in the mixture should be 20% of the entire mixture. So, 0.5 should be 20% of (x + 0.5). In equation form this can be written as:

0.5 = 20% of (x + 0.5)

0.5 = 0.2(x + 0.5)

Dividing both sides by 0.2, we get:

2.5 = x + 0.5

x = 2 lbs

This means, 2 lbs of water should be added to make 20% syrup solution. So, total syrup she would have will be 2.5 lbs.

Case 2: 75% syrup

75% syrup means the sugar percentage in the mixture should be 75% of the entire mixture. So, 0.5 should be 75% of (x + 0.5). In equation form this can be written as:

0.5 = 75% of (x + 0.5)

0.5 = 0.75(x + 0.5)

Dividing both sides by 0.75, we get:

0.67 = x + 0.5

x = 0.17 lbs

This means, 0.17 lbs of water should be added to make 75% syrup solution. So, total syrup she would have will be 0.67 lbs.

Case 3: 5% syrup

5% syrup means the sugar percentage in the mixture should be 5% of the entire mixture. So, 0.5 should be 5% of (x + 0.5). In equation form this can be written as:

0.5 = 5% of (x + 0.5)

0.5 = 0.05(x + 0.5)

Dividing both sides by 0.05, we get:

10 = x + 0.5

x = 9.5 lbs

This means, 9.5 lbs of water should be added to make 5% syrup solution. So, total syrup she would have will be 10 lbs.

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3 years ago