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melamori03 [73]
3 years ago
11

1/2 +3/4 simplest form ??

Mathematics
1 answer:
Feliz [49]3 years ago
6 0
1/2 + 3/4 

1/2 = 2/4 therefore

2/4 + 3/4 = 5/4

5/4 (improper fraction) = 1 1/4 (mixed fraction)

hope this helps
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What are the domain and range of this function y=(x+3)^2 -5
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Note: the range should be [-5, \infty). See explanation below.

===================================================

We can plug in any real number for x to get some output for y. The domain is the set of all real numbers in which we say (-\infty, \infty) which is interval notation. It represents the interval from negative infinity to positive infinity.

The range is the set of possible outputs. The smallest output possible is y = -5 which occurs at the vertex (3,-5). We can get this y value or larger. So we can describe the range as the set of y values such that y \ge -5 and that translates to the interval notation [-5, \infty).

The square bracket says "include this endpoint" while the curved parenthesis says to exclude the endpoint. Your teacher mistakenly wrote (-5, \infty) for choice A, when they should have written [-5, \infty)

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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.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

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

#SPJ4

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