To solve the problem, we will substitute t = 60 into the given equation M(t) = 150 ln(32/(76-t)).

(a) M(60):
M(60) = 150 ln(32/(76-60))
= 150 ln(32/16)
= 150 ln(2)
≈ 207.9

The answer M(60) represents the number of minutes it takes to raise the temperature of an ice cube to 60 degrees Fahrenheit in a room where the air temperature is 76 degrees Fahrenheit. In this case, it takes approximately 207.9 minutes.

(b) To find M'(60), we need to take the derivative of M(t) with respect to t:

M(t) = 150 ln(32/(76-t))
M'(t) = (150) * (d/dt) ln(32/(76-t))

Using the chain rule, (d/dt) ln(32/(76-t)) = (d/dx) ln(x) * (d/dt) (32/(76-t)), where x = 32/(76-t).
(d/dx) ln(x) = 1/x
(d/dt) (32/(76-t)) = (d/dx) (32/(76-t)) * (d/dt) (76-t)

Let's find these derivatives:
(d/dx) ln(x) = 1/x
(d/dt) (32/(76-t)) = (d/dx) (32/(76-t)) * (d/dt) (76-t) = (-32/(76-t)^2) * (-1) = 32/(76-t)^2

Substituting these derivatives back into M'(t):

M'(t) = (150) * (1/x) * (32/(76-t)^2)
= 4800/(x(76-t)^2)

Now, we can find M'(60) by substituting t = 60:

M'(60) = 4800/(32/(76-60)^2)
= 4800/(32/16^2)
= 4800/(32/256)
= 4800/(8)
= 600

The answer M'(60) represents the rate at which the number of minutes changes with respect to the temperature when the temperature is 60 degrees Fahrenheit. In this case, the rate of change is 600 minutes per degree Fahrenheit.

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