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sp2606 [1]
3 years ago
13

If marie wants to decorate a gift box with ribbon. She measures three pieces that she has with a ruler.They are 32. Centimeters,

41.19 centimeters and 57.8 centimeters long.Marie needs 200 centimeters of ribbon for the box. Does she have enough?
Mathematics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer: No, she does not have enough.

Step-by-step explanation:

1. You have the following information given in the problem above:

- The measures of the three pieces are: 32 centimeters, 41.19 centimeters and 57.8 centimeters long.

- She need 200 centimeters of ribbon for the box.

2. Therefore, you must add the measures given in the problem to know if Marie has enough ribbon to decorate the gift box. Then:

32cm+41.19cm+57.8cm=130.99cm

3. As you can see:

 130.99 cm<200 cm

Therefore, she does not have enough ribbon.

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A.  .325L

Step-by-step explanation:

All you have to do is had .125 with .20 to get your answer

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Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
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The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
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Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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Pleaseeee i need help asap
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That's just too bad. I don't the answer. I'm just doing this to get points. Good luck.
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