Question

(t) = 9400(0.4)^(t/60) represents the change in a quantity over t minutes. what ant 0.4 reveal about the rate of change of the quantity? second 60 minutes day 60 seconds tempt 1 out of 1 tion is decaying exponentially at a rate of 60 % every

Explanation:

Step1: Identify the exponential decay formula

The general form of an exponential decay function is \( A(t) = A_0b^{\frac{t}{k}} \), where \( A_0 \) is the initial amount, \( b \) is the decay factor, \( t \) is time, and \( k \) is the time it takes for the decay factor to be applied. If \( b < 1 \), it represents decay, and the decay rate \( r \) is given by \( r=1 - b \).

Step2: Calculate the decay rate

Given the function \( f(t)=9400(0.4)^{\frac{t}{60}} \), here \( b = 0.4 \). The decay rate \( r=1 - 0.4=0.6 \) or \( 60\% \).

Step3: Analyze the time interval

The exponent is \( \frac{t}{60} \), which means that the time unit for the decay is 60 units of time (in this case, since \( t \) is in minutes, the time interval for the decay rate of \( 60\% \) is 60 minutes? Wait, no, wait. Wait, let's re - examine. Wait, the exponent is \( \frac{t}{60} \), so when \( t = 60 \) minutes, the exponent becomes 1, and the function becomes \( f(60)=9400(0.4)^{1}=9400\times0.4 \), which means that over 60 minutes, the quantity is multiplied by 0.4, so the decay rate over 60 minutes? Wait, no, the decay rate is \( 1 - 0.4 = 0.6=60\% \) decay per 60 minutes? Wait, no, let's correct.

Wait, the formula \( A(t)=A_0b^{\frac{t}{k}} \), the time period for the decay factor \( b \) is \( k \). So if \( b = 0.4 \), that means that after a time period of \( k = 60 \) minutes, the quantity is \( 0.4 \) times the original. So the amount of decay is \( 1 - 0.4=0.6 = 60\% \) per 60 minutes. Wait, but the question's dropdown has options like second, 60 minutes, day, 60 seconds. Wait, the exponent is \( \frac{t}{60} \), so \( t \) is in minutes, and the denominator is 60, so the time interval for the decay rate of \( 60\% \) is 60 minutes? Wait, no, maybe I made a mistake. Wait, let's think again.

Wait, the function is \( f(t)=9400(0.4)^{\frac{t}{60}} \). Let's take \( t = 60 \) minutes. Then \( f(60)=9400\times(0.4)^{\frac{60}{60}}=9400\times0.4 \). So in 60 minutes, the quantity is 40% of the original, which means it has decayed by \( 1 - 0.4 = 0.6=60\% \) in 60 minutes. But the dropdown options: the last part of the sentence is "exponentially at a rate of 60% every [dropdown]". The options are second, 60 minutes, day, 60 seconds.

Wait, the exponent is \( \frac{t}{60} \), so when \( t = 60 \) (the unit of \( t \) is minutes), so the time period for the 60% decay is 60 minutes? Wait, no, maybe the unit of \( t \) is seconds? Wait, the problem's context: the original function is about \( t \) minutes? Wait, the problem says "over \( t \) minutes". So \( t \) is in minutes, and the exponent is \( \frac{t}{60} \), so when \( t = 60 \) minutes, the exponent is 1. So the decay of 60% (since \( 1 - 0.4 = 0.6 \)) happens every 60 minutes? But the dropdown has "60 minutes" as an option. Wait, but let's check the options again. The user's input has a partially filled sentence: "tion is decaying exponentially at a rate of 60% every [dropdown]". The dropdown options are second, 60 minutes, day, 60 seconds.

Wait, let's re - express the function. Let's let \( t \) be in seconds. Wait, no, the problem says "over \( t \) minutes". So \( t \) is minutes. The exponent is \( \frac{t}{60} \), so if \( t = 60 \) minutes, the exponent is 1. So the decay rate of 60% (because \( 1 - 0.4=0.6 \)) occurs every 60 minutes? But that seems odd. Wait, maybe the unit of \( t \) is seconds? Wait, no, the problem says "over \( t \) minutes".

Wait, maybe I messed up. Let's start over. The exponential decay formula can also be written as \( A(t)=A_0(1 - r)^{\frac{t}{k}} \), where \( r \) is the decay rate and \( k \) is the time for the decay. Here, \( (1 - r)=0.4 \), so \( r = 0.6 = 60\% \), and \( \frac{t}{k}=\frac{t}{60} \), so \( k = 60 \). So the time period \( k \) is 60 minutes? Wait, no, if \( t \) is in minutes, then \( k = 60 \) minutes, meaning that every 60 minutes, the quantity decays by 60%? But that would mean that after 60 minutes, the quantity is 40% of the original. But let's check the options. The options for the time unit are second, 60 minutes, day, 60 seconds.

Wait, maybe the exponent is \( \frac{t}{60} \) where \( t \) is in seconds? Wait, the problem is a bit unclear, but given the options, and the exponent \( \frac{t}{60} \), the most probable time unit for the 60% decay rate is 60 minutes? Wait, no, wait, 60 seconds is 1 minute. Wait, maybe the exponent is \( \frac{t}{60} \) where \( t \) is in seconds, so that when \( t = 60 \) seconds (1 minute), the exponent is 1. But the problem says \( t \) is in minutes. This is a bit confusing. But given the function \( f(t)=9400(0.4)^{\frac{t}{60}} \), and the decay rate is \( 60\% \) (since \( 1 - 0.4 = 0.6 \)), and the exponent is \( \frac{t}{60} \), the time interval for the decay rate of \( 60\% \) is 60 minutes? Wait, no, let's take an example. Let \( t = 0 \), \( f(0)=9400 \). Let \( t = 60 \) minutes, \( f(60)=9400\times(0.4)^{1}=9400\times0.4 = 3760 \). The amount of decay is \( 9400 - 3760 = 5640 \), and \( \frac{5640}{9400}=0.6 = 60\% \). So over 60 minutes, the quantity decays by 60%. So the answer should be that the quantity is decaying exponentially at a rate of 60% every 60 minutes.

Answer:

The quantity is decaying exponentially at a rate of 60% every 60 minutes. So the answer for the dropdown (the time unit) is 60 minutes.