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3 years ago

What is 5x23x2 solve with properties

Mathematics

1 answer:

Mariulka [41]3 years ago

6 0

I think 5x23x2=23x5x2=23x10=230 the property is the associativity

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15 POINTS!!

Alexxx [7]

X^2 - 10x + 8 =0
x^2 - 10 + (-10/2)^2 - (-10/2)^2 + 8 = 0
(x - 5)^2 - 25 + 8 = 0
(x - 5)^2 - 17 = 0
To be honest , this is the final step for this equation. It seems like there is no any suitable answer for this question..
To me , I think the best answer will be the third option.
x - 5 =0
x = 5

2x-10 = 0
2x = 10
x = 5
I guess this answer seems like legit.. So I will choose the third option.

7 0

3 years ago

Read 2 more answers

50 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

schepotkina [342]

Answer: 17mm

Step-by-step explanation:

$c\sqrt{15^{2}+8^{2} } =17$

6 0

2 years ago

URGENT!! I REALLY NEED HELP

Ivenika [448]

Answer

D is the answer

Step-by-step explanation:

as I can see in the screen shot you don't need explanation lol

4 0

3 years ago

In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any

Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

$E \cap F = \emptyset,\quad E \cup F = \Omega$

where $\Omega$ is the whole sample space.

This implies that

$P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)$

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day $d_1$ .

The second footballer can be born on any other day, so he has 364 possible birthdays:

$d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}$

the probability for the first two footballers to be born on two different days is thus

$1 \cdot \dfrac{364}{365} = \dfrac{364}{365}$

Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

$d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}$

so, the probability for the first three footballers to be born on three different days is

$1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}$

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

$\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

So, the probability that at least two footballers are born on the same day is

$1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}$

since the two events are complementary.

8 0

2 years ago

3 + 4 = ? וא 2/3+2/4

Elina [12.6K]

Answer:

1. 7

2. 17/12 can be a mixed number

8 0

2 years ago