<span>Jill usó piedras que eran cada una 3) 4 de un pie para construir una pared. Dhe apiló 6 piedras en la parte superior de cada pared fraccional</span>
Answer:
Step-by-step explanation:
<u>Given </u><u>:</u><u>-</u><u> </u>
And we need to find out the function from the given options which is equivalent to the given function . For that firstly simplify the whole square term , that is ,
<u>Using </u><u>Identity</u><u> </u><u>:</u><u>-</u>
![\implies (a-b)^2= a^2-2ab + b^2](https://tex.z-dn.net/?f=%5Cimplies%20%28a-b%29%5E2%3D%20a%5E2-2ab%20%2B%20b%5E2)
<u>Whole</u><u> </u><u>square</u><u> </u><u>term </u><u>:</u><u>-</u><u> </u>
![\implies (x-3)^2= x^2+9-6x](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%5E2%3D%20x%5E2%2B9-6x%20)
Now multiply the constant term outside the bracket to it . As ,
Add the constant terms .
<u>H</u><u>ence </u><u>the </u><u>second</u><u> </u><u>option</u><u> is</u><u> </u><u>correct</u><u> </u><u>.</u>
Answer:
or ![\frac{9}{78}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B78%7D)
(see explanation)
Step-by-step explanation:
This problem is asking about compound probability, what is the probability of this even happening three times in a row.
Since there are 20 boy's names and 20 girl's names in the hat, one can say that half of the names in the hat. Assume that after drawing it from the hat, the name is put back into the hat. Therefore, on each night, there is a half-chance of the event happening. To find the probability of the event, one must multiply (
) by itself (3) times,
![(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})\\=\frac{1}{8}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%29%28%5Cfrac%7B1%7D%7B2%7D%29%28%5Cfrac%7B1%7D%7B2%7D%29%5C%5C%3D%5Cfrac%7B1%7D%7B8%7D)
In this case, assume that after each drawing the name is not replaced into the hat. This means that after every draw, one must subtract (1) from both the numerator and denominator of the fraction when the fraction is in the un-simplified form.
![(\frac{1}{2})(\frac{19}{39})(\frac{18}{38})](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%29%28%5Cfrac%7B19%7D%7B39%7D%29%28%5Cfrac%7B18%7D%7B38%7D%29)
Simplify,
![(\frac{1}{2})(\frac{19}{39})(\frac{9}{19})](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%29%28%5Cfrac%7B19%7D%7B39%7D%29%28%5Cfrac%7B9%7D%7B19%7D%29)
![\frac{9}{78}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B78%7D)
9x10 to the -8th power.
Since the numbers are both x10^-8, you can simply add 2 and 7. If you would rather look at it as:
0.00000007 + 0.00000002 = 0.00000009 = 10x10^-8
then you will get the same answer.