Answer:
c. 0.80
Step-by-step explanation:
If two events A and B are mutually exclusive they cannot occur simultaneously, but they can occur independently and so their individual probabilities can be added.
![P(A \cap B) = 0\\P(A \cup B) = P(A) + P(B)](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200%5C%5CP%28A%20%5Ccup%20B%29%20%3D%20P%28A%29%20%2B%20P%28B%29)
Therefore: P(A ∪ B) is:
![P(A \cup B) = 0.3+0.5\\P(A \cup B) = 0.8](https://tex.z-dn.net/?f=P%28A%20%5Ccup%20B%29%20%3D%200.3%2B0.5%5C%5CP%28A%20%5Ccup%20B%29%20%3D%200.8)
Thus, the answer is c. 080
It is (3,5)
I did the Quiz and got a 100%
Answer:
you need 100ml of 5X TAE and 400ml of water.
Step-by-step explanation:
You need to use a rule of three:
![C_1V_1=C_2V_2](https://tex.z-dn.net/?f=C_1V_1%3DC_2V_2)
where:
![\left \{ {{C_1= 5X} \atop {C_2=1X}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7BC_1%3D%205X%7D%20%5Catop%20%7BC_2%3D1X%7D%7D%20%5Cright.)
and
![\left \{ {{V_1 = V_{TAE}} \atop {V_2=500ml}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7BV_1%20%3D%20V_%7BTAE%7D%7D%20%5Catop%20%7BV_2%3D500ml%7D%7D%20%5Cright.)
Therefore:
![V_{TAE} = \frac{V_2*C_2}{C_1}](https://tex.z-dn.net/?f=V_%7BTAE%7D%20%3D%20%5Cfrac%7BV_2%2AC_2%7D%7BC_1%7D)
![V_{TAE} = 100ml](https://tex.z-dn.net/?f=V_%7BTAE%7D%20%3D%20100ml)
Then just rest the TAE volume to the final Volume and you get the amount of water that you need to reduce the concentration.
The answer would be x is less than -6 because you’d divide both sides by -5 and then flip the inequality sign (it’s greater than but flip it to less than) then you’d have x is less than -6 :) hope this helps!