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cricket20 [7]
3 years ago
5

Give an example of an addition problem in which you would and would not group the addenda differently to add

Mathematics
1 answer:
Ann [662]3 years ago
8 0
Pemdas if you don't know what they are it looks like this.. (4+5)-5+4
similar to that..... 
I am pretty sure you wouldn't add that in a agenda because of the different properties it has but there are plenty more ways i just haven't figured them out 

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Nine tiles are numbered $\color[rgb]{0.35,0.35,0.35}1, 2, 3, \ldots, 9$. Each of three players randomly selects and keeps three
Eduardwww [97]

The probability that all three players obtain an odd sum is 3/14.

<h3>What is probability?</h3>

The probability is the ratio of possible distributions to the total distributions.

I.e.,

Probability = (possible distributions)/(total distributions)

<h3>Calculation:</h3>

Given that,

There are nine tiles - 1, 2, 3,...9, respectively.

A player must have an odd number of odd tiles to get an odd sum. That means he can either have three odd tiles, or two even tiles and an odd tile.

In the given nine tiles the number of odd tiles = 5 and the number of even tiles = 4.

The only possibility is that one player gets 3 odd tiles and the other two players get 2 even tiles and 1 odd tile.

So,

One player can be selected in ^3C_1  ways.

The 3 odd tiles out of 5 can be selected in ^5C_3 ways.

The remaining 2 odd tiles can be selected and distributed in ^2C_1 ways.

The remaining 4 even tiles can be equally distributed in \frac{4 ! \cdot 2 !}{(2 !)^{2} \cdot 2 !} ways.

So, the possible distributions = ^3C_1 × ^5C_3 × ^2C_1 × \frac{4 ! \cdot 2 !}{(2 !)^{2} \cdot 2 !}

⇒ 3 × 10 × 2 × 6 = 360

To find the total distributions,

The first player needs 3 tiles from the 9 tiles in ^9C3=84 ways

The second player needs 3 tiles from the remaining 6 tiles in ^6C_3=20 ways

The third player takes the remaining tiles in 1 way.

So, the total distributions = 84 × 20 × 1 = 1680

Therefore, the required probability = (possible distributions)/(total distributions)

⇒ Probability = 360/1680 = 3/14.

So, the required probability for the three players to obtain an odd sum is 3/14.

Learn more about the probability of distributions here:

brainly.com/question/2500166

#SPJ4

3 0
1 year ago
If 70% of college students say they have a job what is the probability that 3 randomly selected college students say they have a
dsp73

Answer: 0.343

Step-by-step explanation:

Probability of college students that say they have a job = 70%

Number of randomly selected college students = 3

The probability that 3 randomly selected college students say they have a job will be:

= (70/100)³

= 0.7 × 0.7 × 0.7

= 0.343

4 0
3 years ago
Find the slope of the line?
LenKa [72]

Answer:

-3/4

Step-by-step explanation:

Option 1

Direction of the line: Down

-Negative Slope

Rise=-3

Run=4

Rise/Run=-3/4

Option 2

Points: (1,-2) and (-3,1)

y2-y1/x2-x1= 1-(-2)/-3-1

1+2/-3-1

-3/4

3 0
3 years ago
Read 2 more answers
Why aren't theories considered absolute truths?
Whitepunk [10]
They are not absolutely truths because they are not found to be truth of false they are things that are only a thing carried over they years and not know if they are the truth or if they are false
6 0
3 years ago
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What is 5/7 + 3/7??????
vaieri [72.5K]
8/7 as an improper and 1 and 1/7 for proper.
6 0
3 years ago
Read 2 more answers
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