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sladkih [1.3K]
3 years ago
10

How are the commutative property of addition and the commutative property of multiplication alike?

Mathematics
2 answers:
swat323 years ago
5 0
Because the sum and product will always have to flipped
Example
4x5=5x4
9+6=6+9
lyudmila [28]3 years ago
4 0

Answer:

The commutative property of addition and the commutative property of multiplication are basically the same. Both state that you can change the order...

Step-by-step explanation:

Commutative Property of Addition: 3+2=2+2

Commutative Property of Multiplication= 3*2=2*3

<em>Hope this helps :)</em>

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Step-by-step explanation:

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Allison has 4 1/2 cups of sugar. She used 2/3 of it to make cookies. How much sugar did Allison use for the cookies?
solmaris [256]
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1 1/2 cups

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An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same
elena-s [515]

Answer:

Approximately 0.15 (360 / 2401.) (Assume that the choices of the 5 passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all 7 floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these 5 passenger could choose from any one of the 7 floors. There would be a total of 7 \times 7 \times 7 \times 7 \times 7 = 7^{5} unique ways for these 5\! passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the 7 floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only (7 - 1) = 6 floors.

Likewise, the third passenger would have to choose from only (7 - 2) = 5 floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only (7 \times 6 \times 5 \times 4 \times 3) unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the 7 floors. Each of these 7^{5} combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}.

5 0
2 years ago
For some integer n, the first, the third and the fifth terms of an arithmetic sequence are respectively 3n, 5n – 6, and 11n + 8.
Mazyrski [523]

Answer:

a₄=8n+1= -39.

Step-by-step explanation:

1) if a₁=3n; a₃=5n-6 and a₅=11n+8, then it is possible to calculate the difference according to 0.5(a₅-a₃)=0.5(a₃-a₁). Then

2) 0.5(11n+8-5n+6)=0.5(5n-6-3n); ⇔ 6n+14=2n-6; ⇔ n= -5.

3) if n=-5, then the 4th term is:

a_4=\frac{a_5+a_3}{2}; \ = > \ a_4=\frac{11n+8+5n-6}{2}=8n+1;

or a₄=-39.

8 0
2 years ago
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