attempt 1: 10 attempts remaining. e(t) gives the proportion (percent written as a decimal) of those living…

attempt 1: 10 attempts remaining. e(t) gives the proportion (percent written as a decimal) of those living in the u.s. who exercise regularly, t years since 2000. p(t) million people gives the u.s. population, t years since 2000. e(6)=0.26, p(6)=304.19, de/dt|t = 6=0.03, dp/dt|t = 6=3.68 complete the following statements: in 2006, the number of people living in the u.s. who exercised regularly was million people. in 2006, the number of people living in the u.s. who exercised regularly was increasing by million people per year.
Answer
Explanation:
Step1: Find number of regular - exercisers in 2006
The number of people who exercise regularly is the product of the proportion of regular - exercisers and the total population. In 2006 ($t = 6$), we use the formula $N=E(6)\times P(6)$. $N = 0.26\times304.19$ $N=79.0894$
Step2: Find the rate of increase of regular - exercisers in 2006
We use the product rule for differentiation. If $N(t)=E(t)P(t)$, then $N^\prime(t)=E^\prime(t)P(t)+E(t)P^\prime(t)$. At $t = 6$, we substitute the given values: $E(6) = 0.26$, $P(6)=304.19$, $E^\prime(6)=0.03$, and $P^\prime(6)=3.68$. $N^\prime(6)=0.03\times304.19+0.26\times3.68$ $N^\prime(6)=9.1257 + 0.9568$ $N^\prime(6)=10.0825$
Answer:
In 2006, the number of people living in the U.S. who exercised regularly was $79.0894$ million people. In 2006, the number of people living in the U.S. who exercised regularly was increasing by $10.0825$ million people per year.