MATH SOLVE

9 months ago

Q:
# In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, the average number of computers in a home could be approximated by q = 0.3458 ln x − 3.045 10,000 ≤ x ≤ 125,000 where x is mean household income. A certain community had a mean income of $30,000, increasing at a rate of $1,000 per year. a. How many computers per household were there? (Round your answer to four decimal places.) 0.5198 Correct: Your answer is correct. computers per household. b. How fast was the number of computers in a home increasing? (Round your answer to four decimal places.) 0.0115 Correct: Your answer is correct. computers per household per year.

Accepted Solution

A:

Answer:a) 0.5198 computers per householdb) 0.01153 computersStep-by-step explanation:Given:number of computers in a home, q = 0.3458 ln x - 3.045 ; 10,000 ≤ x ≤ 125,000here x is mean household incomemean income = $30,000increasing rate, [tex]\frac{dx}{dt}[/tex] = $1,000Now,a) computers per household aresince,mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000thus,q = 0.3458 ln(30,000) - 3.045orq = 0.5198 computers per householdb) Rate of increase in computers i.e [tex]\frac{dq}{dt}[/tex][tex]\frac{dq}{dt}[/tex] = [tex]\frac{d(0.3458 ln x - 3.045)}{dt}[/tex]or[tex]\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0[/tex]on substituting the values, we get[tex]\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000[/tex]or= 0.01153 computers