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3 years ago

Evaluate 6+2•3-1 A) 23 B)16 C)10 D)11

Mathematics

2 answers:

Brums [2.3K]3 years ago

8 0

Answer:

D) 11

Step-by-step explanation:

You always Multiply/Divide first, so you would first multiply 2*3, which equals 6.
after you multiply, you add. 6 + 6 = 12.
Finally, subtract. 12-1 = 11.

If it helps, remember the order of opperations:

PEMDAS.

P = Parenthesis (you don't have any ( ) in this equation, so skip it.)

E = Exponents ( skip this step too. )

M = Multiply, (this is your first step.)

D = Divide

A = Add

S = Subtract

Ratling [72]3 years ago

7 0

6 + 2 * 3 - 1

2 * 3 = 6

6 + 6 - 1

6 + 6 = 12

12 - 1 = 11

D) 11

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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:

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1 year ago

I WILL GIVE 13 POINTS IF YOU SHOW YOUR WORK WITH THE ANSWER A clothing store purchased jeans from the manufacturer at a cost of

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2 years ago

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She bought <br> 314yards of fabric. Her total cost was $13. What was the cost per yard?

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Answer:

$24.15 (answer is rounded)

Step-by-step explanation:

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