All categories

2 years ago

HELP i need it i will give brainly

Mathematics

2 answers:

sergiy2304 [10]2 years ago

6 0

Answer:

2/3^15

Step-by-step explanation:

9^-8 * 3/2^-1

= 3^-16 * 3/2^-1

= 3^-15/2^-1

= 1/2^-1 * 3^15

= 2/3^15

iogann1982 [59]2 years ago

5 0

Answer:

2/9^8

Step-by-step explanation:

3^0 = 1 so it turns into 9^-8/ 2^-1

Then put 9^-8 on the bottom to make it positive

Now its 1/9^8 & 2^-1

Then put 2 on top to make it positive and 2^1 is just 2

So now you have 2/9^8

You might be interested in

ANSWER ASAP PLEASE .

ankoles [38]

Answer:

a,d,e and f.

Step-by-step explanation:

hope this helps!

please mark me as brainliest. Thanks!

8 0

3 years ago

You deposit $300 in a savings account. The account earns 2% simple interest per year.

Anit [1.1K]

300x .02= 6

6x10=60

Therefore the answer is $360

6 0

3 years ago

Whats a square + b square I don't know thats why I am asking u dude

murzikaleks [220]

Answer:

.............................

8 0

3 years ago

ABCEDEFGH is a regular octagon. The minimum degree of rotation by which this octagon can map onto itself is . It will take incre

Mrrafil [7]

Answer:

45 or multiples of 45 around the centre of octagon

Step-by-step explanation:

Given that ABCDEFGH is a regular octagon. i.e. it has 8 sides.

Since regular ocagon, all interior angles would be equal.

Sum of all interior angles of octagon = 2(8)-4 right angles

= 12 right angles

Hence each angle = 12(90)/8 = 135 degrees

Thus the octagon when rotated will take the same shape if vertices interchange also due to the property that all sides and angles are equal

Since each angle is 135 imagine an octagon with one vertex at origin O, and adjacent vertex B on x axis. OB has to be coincident with BC the next side or the previous side to get it mapped onto itself

The centre will be at the middle with each side subtending an angle of 45 degrees.

Hence if rotation is done around the centre with 45 degrees we will get octagon mapped onto itself.

45, 90, 135 thus multiples of 45

8 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

3 years ago