Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
Answer:
Assignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques
Step-by-step explanation:
Assignment: 01.07 LaboratoryAssignment: 01.07 Laboratory Techniques TechniquesAssignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if a new feed + additive generates a better production of milk in cows. For this, the owner selects 13 cows and randomly separates them into two groups.
Group 1 has 8 cows that receive the new feed + additive.
Group 2 has 5 cows that were fed with the old feed.
After two weeks of feeding the animals with the different feeds, the production of milk of each group was recorded so that they can be compared.
Since you have two separate groups to wich at random two different treatments were applied and later the variable was measured, these two samples/groups are independent and the proper test to compare the population means of the milk production in both groups is a pooled t.
I hope this helps!
Answer:
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Step-by-step explanation:
"half of R" it is: 
"y cubed" it is: 
Therefore, half of R subtracted from y cubed is: 
