A cup of coffee at 93 degrees Celsius is placed in a room at 23 degrees Celsius. Suppose that the coffee cools at a rate of 1 degrees Celsius per minute when the temperature of the coffee is 70 degrees. The differential equation describing this has the form dTdt=k(T−A), where A= 23 , and k= -1 . Either from drawing the direction field (try Maple or by hand) or other intuitive means, the limiting value of T should be:

Step-by-step explanation:

The rate of change of temperature of the coffee with respect to time is expressed by the differential equation

dT/dt=k(T−A) when k = -1 and A = 23. The equation will become:

dT/dt = -(T-23)

If the coffee cools at the rate of 1°C per minute, then dT/dt = -1 and T = 70

Substituting into the equation:

-1 = k(70-23)

-1 = 47k

k = -1/47

k = -0.0213

Substituting k = -0.0213 into the original equation, the differential equation will be:

dT/dt=k(T−A)

dT/dt = -0.0213(T-23)

dT/dt = -0.0213T+0.489

To get the value of T, we will use variable separable method

dt = dT/-0.0213T+0.489

Integrate both sides

t = -0.0213ln(-0.0213T+0.489)

At t = 1 minute

1 = -0.0213ln(-0.0213T+0.489)

1/-0.0213 = ln(-0.0213T+0.489)

-46.95 = ln(-0.0213T+0.489)

Apply exp to both sides

e^-46.95 = e^ln(-0.0213T+0.489)

4.073×10^-21= -0.0213T+0.489

-0.0213T = 4.073×10^-21-0.489

-0.0213T = -0.489

T = 0.489/0.0213

T = 22.96°C

two data sets with the same median will not have the same varaibility

step-by-step explanation: