Question

= 9500(0.55)^{\frac{t}{k}} represents the change in a quantity over t minutes. what 0.55 reveal about the rate of change of the quantity
60 seconds
60 minutes
day
second
decreasing exponentially at a rate of % every

Explanation:

Step1: Recall exponential decay formula

The general form of an exponential decay function is \( Q(t) = Q_0(1 - r)^{\frac{t}{T}} \) or \( Q(t)=Q_0a^{\frac{t}{T}} \), where \( Q_0 \) is the initial quantity, \( r \) is the decay rate, \( T \) is the time period for the decay factor, and \( a = 1 - r \) (since \( 0 < a< 1 \) for decay). In the given function \( Q(t)=9500(0.55)^{\frac{t}{60}} \) (assuming the exponent is \( \frac{t}{60} \) from the "60 seconds" option context), we compare with the form \( Q(t)=Q_0a^{\frac{t}{T}} \). Here, \( a = 0.55 \), so the decay factor per \( T \) time is \( 0.55 \), meaning the remaining quantity is \( 55\% \) of the previous, so the decay rate \( r=1 - 0.55 = 0.45 \) or \( 45\% \) loss, but the base \( 0.55 \) implies that for each time period \( T \) (here, looking at the exponent \( \frac{t}{60} \), so \( T = 60 \) seconds), the quantity is multiplied by \( 0.55 \). So the time period for the decay factor (the "every" time) is 60 seconds, and the rate of change: since \( 0.55<1 \), it's decreasing exponentially, with the base \( 0.55 \) meaning the quantity is \( 55\% \) of the previous after 60 seconds, so the time unit is 60 seconds.

Step2: Identify the time unit from the exponent

The exponent is \( \frac{t}{60} \), which means \( t \) is in seconds, and we're dividing by 60, so the time period for the exponential change (the "every" part) is 60 seconds. The base \( 0.55 \) shows that for each 60 - second interval, the quantity is multiplied by \( 0.55 \), so it's decreasing exponentially at a rate where each 60 seconds, the quantity is \( 55\% \) of what it was, so the time unit is 60 seconds.

Answer:

60 seconds