Solve the problem.
-Instruments on a satellite measure the amount of power generated by the satellite's power supply. The time t and the power P can be modeled by the funct $$P = 50 e ^ { - t / 300 }$$ , where t is in days and P is in watts. How much power will be available after 378 days? Round to the nearest hundredth.
A)14.32 watts
B)4.11 watts
C)14.18 watts
D)15.95 watts

Question

Quizplus · Accepted Answer

To find the power available after 378 days, substitute t = 378 into the function:  $$ P = 50 e^{-378/300} $$  $$ P = 50 e^{-1.26} $$  $$ P \approx 50 \times 0.283 $$  $$ P \approx 14.18 $$ watts.

Solve the Problem P=50e−t/300P = 50 e ^ { - t / 300 }P=50e−t/300

Solve the Problem $P = 50 e ^ { - t / 300 }$