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Harman [31]
3 years ago
8

How do you divide 2.16 by 3

Mathematics
2 answers:
fenix001 [56]3 years ago
5 0

Answer:

The answser is 0.72 because you divide 2.18 by 3

Step-by-step explanation:

for using bace ten blocks you have too move the decimal number 2 spaces to the right. 2.16-->216 You need 21 ten rods and 6 cubes. Why: I wouldn't do 2 hundred flats because 2 isn't divisible by 3. Put the ten rods into 3 equal group, then do the same with the cubes. You should end up with 7 in each group since 21 divided by 3 is 7. You should also end up with 2 cubes in each group. So in one group you have 72. Now move the decimal back to where it came from, 2 spaces to the left. Now you have 0.72.

lozanna [386]3 years ago
3 0

Answer:

by moving the decimal in both numbers twice to the right and doing normal long division (the decimal in 3 is at the end) to get the answer of 0.72

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Inequality

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There are an infinite number of quantities less than 6 (5, 4, 3, 2, 1, 0, -1, -2, -3...) so it is much easier to write an inequality

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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.

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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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