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jenyasd209 [6]
3 years ago
9

The perimeter of a quadrilateral is 38 yards. If three of the sides measure 5.8 yards, 7 yards, and 11.2 yards, what is the leng

th of the fourth side?
Mathematics
1 answer:
Lilit [14]3 years ago
6 0

Answer:

14 yds

Step-by-step explanation:

To find the perimeter, we add up all the sides

P = s1 + s2+ s3 + s4

38 = 5.8+7+11.2 + s4

Combine like terms

38 = 24+s4

Subtract 24 from each side

38-24 = 24-24 +s4

14 = s4

The 4th side is 14 yds

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PLEASE HELP ME!
Troyanec [42]

If 450 is 1/4 of the amount that Henry has, you can find Henry’s amount by multiplying 450 • 4 which gives you an answer of 1800

Or, if you’re looking for a more complicated way, which I doubt, you can set up the problem as:

1/4x=450 and solve by dividing 450 by 1/4

7 0
3 years ago
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
Y=x+1 and y=-x-3 in graphing,Substitution,and elemination
marysya [2.9K]

\left\{\begin{array}{ccc}y=x+1\\y=-x-3\end{array}\right\\\\GRAPHING\ \text{(look at the picture)}\\\\y=x+1\\for\ x=0\to y=0+1=0\to(0,\ 1)\\for\ x=-1\to y=-1+1=0\to(-1,\ 0)\\\\y=-x-3\\for\ x=0\to y=-0-3=-3\to(0,\ -3)\\for\ x=-3\to y=-(-3)-3=0\to(-3,\ 0)\\\\Solution:\ (-2,\ -1)


SUBSTITUTION\\\\\left\{\begin{array}{ccc}y=x+1\\y=-x-3\end{array}\right\\\\\text{substitute}\ y=-x-3\ \text{to the first equation}\\\\-x-3=x+1\ \ \ \ |+3\\-x=x+4\ \ \ \ |-x\\-2x=4\ \ \ \ |:(-2)\\x=-2\\\\\text{substitute the value of x to the second equation}\\\\y=-(-2)-3=2-3=-1\\\\Solution:\ (-2,\ -1)


ELIMINATION\\\\\underline{+\left\{\begin{array}{ccc}y=x+1\\y=-x-3\end{array}\right}\ \ \ |\text{add both sides of the equations}\\.\ \ \ \ 2y=-2\ \ \ \ |:2\\.\ \ \ \ \ \ y=-1\\\\\text{put the value of y to the first equation}\\\\-1=x+1\ \ \ \ |-1\\x=-2\\\\Solution:\ (-2,\ -1)

5 0
3 years ago
I need help plz I don't get this problem
OlgaM077 [116]
To find the answer you need to write equivalent fractions with larger denominators. If you multiply each fraction by 3/3 then your equivalent fractions would be

3/5 × 3/3 = 9/15
4/5 × 3/3 = 12/15

Now we have two fractions that are equivalent to the original fractions and we can easily see that 2 fractions between 9/15 and 12/15 would be...

10/15 and 11/15

(10/15 can reduce to 2/3 so your answers could also be 2/3 and 11/15)
5 0
3 years ago
Use the distributive property <br><br> 5(x-5)=5<br><br><br> I need help asap :(
Inessa [10]
5x-25=5, x=6; multiply the 5 times x and negative 5, then add 25 to both sides, and divide by 5 to get 6.
8 0
3 years ago
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