lesson 10 practice problems\n1. a store receives 2,000 decks of popular trading cards. the number of decks…

lesson 10 practice problems\n1. a store receives 2,000 decks of popular trading cards. the number of decks of cards is a function, d, of the number of days, t, since the shipment arrived. here is a table showing some values of d.\ncalculate the average rate of change for the following intervals:\n a. day 0 to day 5\nb. day 15 to day 20\n t d(t)\n0 2,000\n5 1,283\n10 823\n15 528\n20 338

lesson 10 practice problems\n1. a store receives 2,000 decks of popular trading cards. the number of decks of cards is a function, d, of the number of days, t, since the shipment arrived. here is a table showing some values of d.\ncalculate the average rate of change for the following intervals:\n a. day 0 to day 5\nb. day 15 to day 20\n t d(t)\n0 2,000\n5 1,283\n10 823\n15 528\n20 338

Answer

Explanation:

Step1: Recall average rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Here, $d(t)$ is the function, $t$ is the independent variable.

Step2: Calculate average rate of change for 0 to 5

For the interval from $t = 0$ to $t = 5$, $t_1=0$, $t_2 = 5$, $d(t_1)=2000$, $d(t_2)=1283$. The average rate of change is $\frac{d(5)-d(0)}{5 - 0}=\frac{1283 - 2000}{5}=\frac{-717}{5}=-143.4$.

Step3: Calculate average rate of change for 15 to 20

For the interval from $t = 15$ to $t = 20$, $t_1 = 15$, $t_2=20$, $d(t_1)=528$, $d(t_2)=338$. The average rate of change is $\frac{d(20)-d(15)}{20 - 15}=\frac{338 - 528}{5}=\frac{-190}{5}=-38$.

Answer:

a. - 143.4 b. - 38