Please answer ASAP!<br>Also, please show your thinking!<br>Thank you so much!

Write an equation and use it to solve this question.

HACTEHA [7]

C. five sixth meters

5 0

3 years ago

Read 2 more answers

3/4 + 3/9 how do i do this

Artist 52 [7]

Answer:

13/12

Step-by-step explanation:

Well, you can't add it yet because the denominators don't match. We can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change as well. So, multiply 3 by 8, and get 24. Since 4 is evenly divided by 8, we can multiply just one term to get a common denominator. Multiply 3 by 2, and get 6, then we multiply 4 by 2 and get 8. Since our denominators match, we can add the numerators. So now your answer should be 9/8. Well guess what... IT IS THE ANSWER! You can't reduce it anymore so 9/8 is your answer! Hope this helped!

7 0

3 years ago

Write the expression as the sine or cosine of an angle sin(pi/7) cos(x) + cos(pi/7) sin(x)

Natasha_Volkova [10]

Answer:

$\sin(\frac{\pi}{7}+x)$

Step-by-step explanation:

We are going to use the identity

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

because this identities right hand side matches your expression where

$a=\frac{\pi}{7}$ and $b=x$ .

So we have that $\sin(\frac{\pi}{7})\cos(x)+\cos(\frac{\pi}{7})\sin(x)$ is equal to $\sin(\frac{\pi}{7}+x)$ .

5 0

4 years ago

logan had a bank account balance of 287 he then deposit 35 and went on a shopping trip if he spends 78 what is his new balance a

gizmo_the_mogwai [7]

The answer is 244 because you add 35 to 287 then subtract 78 and it equals 244

6 0

4 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

3 years ago

Please answer ASAP!Also, please show your thinking!Thank you so much!