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4 years ago

Help help help help

Mathematics

2 answers:

sergiy2304 [10]4 years ago

6 0

Try Helena. This is just a wild guess by the way so if i’m wrong dont get mad LOL.

liraira [26]4 years ago

3 0

Answer:

The correct answer is Helena

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Simplify. 4−16÷4+42 −16 13 15 16

Dimas [21]

Answer:

[See Below]

Step-by-step explanation:

✦ First split the equation in little pieces and solve them:

✧ 4 - 16 = -12

✧ 4 + 4² = 20 (4² = 16)

✦ Now divide them:

✧ -12 ÷ 20 = -0.6

So I'm guessing you're trying to simplify 4²... so it'd be 16 because none of the choices are the simplified version of the problem.

~<em>Hope this helps Mate. If you need anything feel free to message me.</em>

7 0

3 years ago

I need help with this one too Please help !!!!!!!!!!! Will mark brainiest

kotykmax [81]

The formula is a^2+b^2=c^2
C is always the hypotenuse so set it up as 4^2+6^2=(3y-2)^2

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

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