A large mixing tank initially contains 1000 gallons of water in which 40 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 5 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 3 pounds of salt per gallon. If img represents the amount of salt in the tank at time t, the correct differential equation for A is:. A) dA/dT=3--.005A
B) dA/dT=5--.05A
C) dA/dT=15--.005A
D) dA/dT=3--.05A
E) dA/dT=15+.05A

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

Step-by-step explanation:

Given the data in the question;

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?

dA/dt = rate in - rate out

first we determine the rate in and rate out;

rate in = 3pound/gallon × 5gallons/min = 15 pound/min

rate out = A pounds/1000gallons × 5gallons/min  = 5Ag/1000pounds/min

= 0.005A pounds/min

so we substitute

dA/dt = rate in - rate out

dA/dt =  15 - 0.005A

Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

yor answer is 7/6 times 5

step-by-step explanation: