Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Sissy imposter among us haha
Answer:
C
Explanation:
the n value must always be greater than the l value
Answer:
class sum (
public static void sumofvalue (int m, int n, int p)
{
System.out.println(m);
System.out.println(n);
System.out.println(p);
int SumValue=m+n+p;
System.out.println("Average="+Sumvalue/3);
}
)
Public class XYZ
(
public static void main(String [] args)
{
sum ob=new sum();
int X=3;
int X=4;
int X=5;
ob.sumofvalue(X,Y,Z);
int X=7;
int X=8;
int X=10;
ob.sumofvalue(X,Y,Z);
}
)
Explanation:
The above program is made in Java, in which first we have printed value in a separate line. After that, the average value of those three values has been printed according to the question.
The processing of the program is given below in detail
* The first one class named 'sum' has been created which contains the function to print individual value and the average of those three values.
* In seconds main class named 'XYZ', the object of that the above class had been created which call the method of the above class to perform functions.
* In the main class values are assigned to variables X, Y, Z.
http://century.rochester.k12.mn.us/cms/One.aspx?portalId=3086882&pageId=6133921
Hope this helps!
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