What is the square root of 49/100?

If B is the midpoint of AC and the length of AC is 23.8 what is the legength of AB

swat32

Hey there,

So explain and to illustrate (how $B$ is the midpoint of $AC$ and the length of $AC$ is 23.8 what is the length of $AB$

<span> A-----------------B----------------C
</span>
So, $\boxed{AC}$ in the problem here is to be considered the (Whole area of this line). $AC$ is the whole thing.

So, what we do in this problem, we do . . .

23.8 ÷ 2 because we need to find what is $AB$ , so we will find the one half of it, which in this illustration is to be considered $AB$

23.8 ÷ 2 = $\boxed{11.9}$

$\boxed{AB=11.9}$

Hope this helps.
~Jurgen

4 0

4 years ago

Find the slope of a line through the given points

Ostrovityanka [42]

Use the formula $\frac{ y_{2} - y_{1} }{ x_{2}- x_{1} }$
In this case: $\frac{-2-4}{1--1}$

You should get -3 when you simplify further.

3 0

3 years ago

I need help solving for x please

Bess [88]

Answer:

x is 55 degrees because the whole angle is a total of 180 degrees. All you have to do is do 180 minus the given angle, 125

8 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

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B

Pachacha [2.7K]

Y=3/4x-2

the y intercept is -2
and the slope is 3/4
slope= change in y/change in x

6 0

3 years ago

What is the square root of 49/100?​

What is the square root of 49/100?