Question

between 12:00 pm and 1:00 pm, cars arrive at a banks drive - thru at the rate of 18 cars per hour (0.3 car per minute). the following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 12:00 pm. complete parts (a)-(c).
f(t)=1 - e^{-0.3t}
(a) determine how many minutes are needed for the probability to reach 30%.
about minutes are needed for the probability to reach 30%.
(round to two decimal places as needed.)

Explanation:

Step1: Set up the equation

We know that $F(t)=0.3$ and $F(t) = 1 - e^{-0.3t}$. So, $0.3=1 - e^{-0.3t}$.

Step2: Rearrange the equation

First, we rewrite the equation as $e^{-0.3t}=1 - 0.3$. Then $e^{-0.3t}=0.7$.

Step3: Take the natural - logarithm of both sides

Taking the natural - logarithm of both sides, we get $\ln(e^{-0.3t})=\ln(0.7)$. Using the property $\ln(e^x)=x$, we have $- 0.3t=\ln(0.7)$.

Step4: Solve for t

Dividing both sides by $-0.3$, we get $t=\frac{\ln(0.7)}{-0.3}$.

Step5: Calculate the value of t

We know that $\ln(0.7)\approx - 0.3567$. Then $t=\frac{-0.3567}{-0.3}\approx1.19$.

Answer:

$1.19$