Answer:
Slope m = 5
Step-by-step explanation: sorry if Im wrong but If i am then it is just 5:)
Answer:
If 10 is the average then 10 each side would be 20
so that 15 fits into 20 at 3/4
1- 3/4 = 1/4
So the answer is 1/4
Step-by-step explanation:
Answer:
True
Step-by-step explanation:
Type I and Type II are not independent of each other - as one increases, the other decreases.
However, increases in N cause both to decrease, since sampling error is reduced.
A small sample size might lead to frequent Type II errors, i.e. it could be that your (alternative) hypotheses are right, but because your sample is so small, you fail to reject the null even though you should.
Given that <span>a bag contains 26 tiles marked with the
letters A through Z.
The probability of picking a letter from the name JACK is 4 / 26
The probability of picking a letter from the name BEN is 3 / 26.
Therefore, the probability of picking a letter from
the name JACK or from the name BEN iis given by 4 / 26 + 3 / 26 = 7 / 26 </span>
Answer:
Step-by-step explanation:
Volume of tank is 3000L.
Mass of salt is 15kg
Input rate of water is 30L/min
dV/dt=30L/min
Let y(t) be the amount of salt at any time
Then,
dy/dt = input rate - output rate.
The input rate is zero since only water is added and not salt solution
Now, output rate.
Concentrate on of the salt in the tank at any time (t) is given as
Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000
dy/dt= dV/dt × dM/dV
dy/dt=30×y/3000
dy/dt=y/100
Applying variable separation to solve the ODE
1/y dy=0.01dt
Integrate both side
∫ 1/y dy = ∫ 0.01dt
In(y)= 0.01t + A, .A is constant
Take exponential of both side
y=exp(0.01t+A)
y=exp(0.01t)exp(A)
exp(A) is another constant let say C
y(t)=Cexp(0.01t)
The initial condition given
At t=0 y=15kg
15=Cexp(0)
Therefore, C=15
Then, the solution becomes
y(t) = 15exp(0.01t)
At any time that is the mass.