The answer would be <span>4*|-2| </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.
Solution: We are given:
Let 
be the weight (oz) of laptop
We have to find 
To find the this probability, we need to find the z score value.
The z score is given below:
Now, we have to find 
Using the standard normal table, we have:
0.9236 or 92.36% of laptops are overweight
Answer:
Step-by-step explanation:
Givens
- The top soil weighs 40 pounds per bag.
- The mulch weighs 20 pounds per bag.
- The cart can only carry up to 480 pounds.
Notice that the restriction is a maximum of 480 pounds, that means the inequality must include the sign 
.
Now, let's call the top soil and 
the mulch, the inequality that represents this problem, would be
Answer: either of two angles whose sum is 180°.
Step-by-step explanation:
