Question

1. which of the following functions represents an exponential decay with a 15% rate of change where a represents the initial value and x represents time in years? a. y = a(0.15)^x b. y = a(0.85)^x c. y = a(1.15)^x d. y = a(1.85)^x 8. a new type of cell phone is being released to the public. the function below is used to predict f(t), the total sales in thousands of units per day based on t, the time in weeks since the date of release. f(t)=40(1.18)^t according to the function, what are the predicted sales on the first day, and what is the predicted rate of percentage increase per week in the sales amount? a. 18,000 units on the first day then 40% per week b. 40,000 units on the first day then 1.18% per week c. 40,000 units on the first day then 18% per week d. 118,000 units on the first day then 40% per week

Explanation:

Step1: Recall exponential - decay formula

The general formula for exponential decay is $y = A(1 - r)^x$, where $A$ is the initial value, $r$ is the rate of decay, and $x$ is the time. Given a 15% rate of decay ($r = 0.15$), the function is $y=A(1 - 0.15)^x=A(0.85)^x$.

Step2: Analyze the sales - function for initial value and growth rate

For the function $f(t)=40(1.18)^t$, when $t = 0$ (first - day, since $t$ is in weeks and at the start $t = 0$), $f(0)=40(1.18)^0=40$. Since $f(t)$ is in thousands of units, the number of units sold on the first day is $40\times1000 = 40000$ units. The general form of an exponential - growth function is $y = a(1 + r)^t$, where $a$ is the initial value and $r$ is the growth rate. In the function $f(t)=40(1.18)^t$, comparing with $y = a(1 + r)^t$, we have $1 + r=1.18$, so $r = 0.18$ or 18% growth rate per week.

Answer:

1. B. $y = A(0.85)^x$
2. C. 40,000 units on the first day then 18% per week