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love history [14]
3 years ago
12

Maria made 171 for 9 hours of work. at the same rate, how much would she make for 7 hours of work

Mathematics
2 answers:
Amanda [17]3 years ago
4 0

Answer: 133

Step-by-step explanation:

171 for 9 hours

171/9 equals 19

So 19 dollars per hour

7 hours times 19 dollars per hour equals 133

Likurg_2 [28]3 years ago
3 0
171 divided by 9 then timed the answer by seven:
= 133
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Please help me with this problem
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i calculated this problem and it's just 46.656

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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

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3 0
2 years ago
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Which expression is equivalent to the product of p+7/3 and 6/p , where p is not equal to 0?
vichka [17]

Answer:

\frac{(6p+14)}{p}

Step-by-step explanation:

(p+\frac{7}{3})(\frac{6}{p})

Making denominator same in first bracket we get

(\frac{3p+7}{3})(\frac{6}{p})

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Dividing 6 by 3 we get 2

(\frac{(3p+7)*2}{p})

using distributive law

(\frac{6p+14}{p})\\

Hence this is our answer

8 0
2 years ago
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