At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):
![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :
![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)
35 because if you half the amount of salads he's eaten (seeing as he eats one burger for every two salads) then the extra one is the first salad in the cycle.
Answer:
You have to use PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction), and solve in that order. Your question is:
2 + 32 ⋅ (3 − 1)
First, solve the Parentheses part of the question. 3-1 is 2. Your question is now:
2+32*2
Now, solve the multiplication part of the question. 32*2 is 64. Your question is now:
2+64
This is now an easy problem to solve.
The answer is 66.
Hope this helps! :)
Answer:
Null hypothesis: ∪ = No possible child abuse or neglect
Alternative hypothesis: Uₐ = Possible child abuse or neglect
Step-by-step explanation:
Null hypothesis: ∪ = No possible child abuse or neglect
Alternative hypothesis: Uₐ = Possible child abuse or neglect
A type I error occurs when you reject the null hypothesis when it is true. In this situation, a type I error occurs when you conclude on possible child neglect or abuse and place the child in protective custody
A type II error occurs when you accept the null hypothesis when it is false. In this instance, a type II error occurs when you conclude on no possible child abuse or neglect when there is and fail to remove the child from the home.
In this case, the type II error is the more serious error. Failure to remove the child when there is possible child abuse or neglect will lead to more detrimental effect. Although, the type I error is also serious, it is not so detrimental as the type II error.
The equation is false so no solution.
Simplify 8/5 * -6 to -48/5
Move the negative sign to the left
Simplify brackets
Since 48/5 = -54 its false so no answer
