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melomori [17]
2 years ago
11

For a area and perimeter math question what is 2/3x2/3????????

Mathematics
1 answer:
Gekata [30.6K]2 years ago
6 0

Answer:

4/9

Step-by-step explanation:

2/3 × 2/3 is 4/9

2×2 is 4 and 3×3 is 9

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Step-by-step explanation:

<h2>Elimination Method</h2>

(x+ 2y=6)2

(2x+4y=12)1

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7 0
3 years ago
-<br>nun<br>THE SUM OF THREE TIMES A<br>Numbers and 11​
Vikentia [17]

Answer:

3x + 11

Step-by-step explanation:

Remember BPEMDAS.

"Three times a number" is saying multiply 3 times a variable; in this case, <em>x</em>. So we have 3x

"The sun of 3x and 11" is saying add our 3x and 11, so: 3x + 11

8 0
4 years ago
A student has a rectangular book. The length of the book is 35
Anna11 [10]

Answer: P = 2(35) + 2(20) = 110 cm

Step-by-step explanation:

2 sides at 35 cm plus 2 sides at 20 cm

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3 0
3 years ago
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Expected Value (50 points)
garri49 [273]

Let W be the random variable representing the winnings you get for playing the game. Then

W=\begin{cases}10-1=9&\text{if the dice sum is odd}\\5-1=4&\text{if the dice sum is 4 or 8}\\50-1=49&\text{if the dice sum is 2 or 12}\\-1&\text{otherwise}\end{cases}

First thing to do is determine the probability of each of the above events. You roll two dice, which offers 6 * 6 = 36 possible outcomes. You find the probability of the above events by dividing the number of ways those events can occur by 36.

  • The sum is odd if one die is even and the other is odd. This can happen 2 * 3 * 3 = 18 ways. (3 ways to roll even with the first die, 3 ways to roll odd for the die, then multiply by 2 to count odd/even rolls)
  • The sum is 4 if you roll (1, 3), (2, 2), or (3, 1), and the sum is 8 if you roll (2, 6), (3, 5), (4, 4), (5, 3), or (6, 2). 8 ways.
  • The sum is 2 if you roll (1, 1), and the sum is 12 if you roll (6, 6). 2 ways.
  • There are 36 total possible rolls, from which you subtract the 18 that yield a sum that is odd and the other 10 listed above, leaving 8 ways to win nothing.

So the probability mass function for this game is

P(W=w)=\begin{cases}\frac12&\text{for }w=9\\\frac29&\text{for }w=4\text{ or }w=-1\\\frac1{18}&\text{for }w=49\\0&\text{otherwise}\end{cases}

The expected value of playing the game is then

E[W]=\displaystyle\sum_ww\,P(W=w)=\frac92+\frac89-\frac29+\frac{49}{18}=\frac{71}9

or about $7.89.

7 0
3 years ago
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