1 12 13 14 15 16 17 18
2 23 24 25 26 27 28
3 34 35 36 37 38
4 45 46 47 48
5 56 57 58
6 67 68
7 78
8
approximately 28 pairs. (breakdown above) hope this helps
Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).
![\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}](https://tex.z-dn.net/?f=%5Cfrac%7BdQ%7D%7Bdt%7D%20%3D%5Ctextrm%20%7Bincoming%20rate%20-%20outgoing%20rate%7D)
Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is =
g/L
Outgoing rate =
![\frac{Q(t)}{10} g/min](https://tex.z-dn.net/?f=%5Cfrac%7BQ%28t%29%7D%7B10%7D%20g%2Fmin)
![\frac{dQ}{dt} = a- \frac{Q(t)}{10}](https://tex.z-dn.net/?f=%5Cfrac%7BdQ%7D%7Bdt%7D%20%3D%20a-%20%5Cfrac%7BQ%28t%29%7D%7B10%7D)
![\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdQ%7D%7B10a-Q%28t%29%7D%20%3D%5Cfrac%7B1%7D%7B10%7D%20dt)
Integrating both sides
![\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cint%20%5Cfrac%7BdQ%7D%7B10a-Q%28t%29%7D%20%3D%5Cint%5Cfrac%7B1%7D%7B10%7D%20dt)
[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram
![\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c](https://tex.z-dn.net/?f=%5CRightarrow%20-log%7C10a-15%7C%3D%5Cfrac%7B1%7D%7B10%7D%5Ctimes%2020%20%2Bc)
![\Rightarrow -log|10a-15|-2=c](https://tex.z-dn.net/?f=%5CRightarrow%20-log%7C10a-15%7C-2%3Dc)
Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get
![- log|10a|= -log|10a-15| -2](https://tex.z-dn.net/?f=-%20log%7C10a%7C%3D%20-log%7C10a-15%7C%20-2)
![\Rightarrow- log|10a|+log|10a-15|= -2](https://tex.z-dn.net/?f=%5CRightarrow-%20log%7C10a%7C%2Blog%7C10a-15%7C%3D%20-2)
![\Rightarrow log|\frac{10a-15}{10a}|= -2](https://tex.z-dn.net/?f=%5CRightarrow%20log%7C%5Cfrac%7B10a-15%7D%7B10a%7D%7C%3D%20-2)
![\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}](https://tex.z-dn.net/?f=%5CRightarrow%20%7C%5Cfrac%7B10a-15%7D%7B10a%7D%7C%3De%20%5E%7B-2%7D)
![\Rightarrow 1-\frac{15}{10a} =e^{-2}](https://tex.z-dn.net/?f=%5CRightarrow%201-%5Cfrac%7B15%7D%7B10a%7D%20%3De%5E%7B-2%7D)
![\Rightarrow \frac{15}{10a} =1-e^{-2}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B15%7D%7B10a%7D%20%3D1-e%5E%7B-2%7D)
![\Rightarrow \frac{3}{2a} =1-e^{-2}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B3%7D%7B2a%7D%20%3D1-e%5E%7B-2%7D)
![\Rightarrow2a= \frac{3}{1-e^{-2}}](https://tex.z-dn.net/?f=%5CRightarrow2a%3D%20%5Cfrac%7B3%7D%7B1-e%5E%7B-2%7D%7D)
![\Rightarrow a = 1.73](https://tex.z-dn.net/?f=%5CRightarrow%20a%20%3D%201.73)
Therefore the concentration of salt in the incoming brine is 1.73 g/L
Well if it is a number like .933333333 you will have to round it, So now it is .93 and .93 in a fraction is 93/100
Answer:
5/11
Step-by-step explanation:
Probability is the chance that something will happen. Basically, it answers the question of how many times an event will occur in a given number of opportunities.
In this example, the word "GEOMETRICAL" has 11 letters and 5 of them are vowels. They are: E, O, E, I, A. The probability of picking a vowel is 5/11. We could say that there are 5 out of 11 chances of a vowel to occur whenever a piece of paper is taken out of the bag.
Answer:
try the answer D
Step-by-step explanation: