Respuesta :
Answer:
a) 0.5198 computers per household
b) 0.01153 computers
Step-by-step explanation:
Given:
number of computers in a home,
q = 0.3458 ln x - 3.045 ; 10,000 ≤ x ≤ 125,000
here x is mean household income
mean income = $30,000
increasing rate, [tex]\frac{dx}{dt}[/tex] = $1,000
Now,
a) computers per household are
since,
mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000
thus,
q = 0.3458 ln(30,000) - 3.045
or
q = 0.5198 computers per household
b) 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
The computers per household are 0.5187 and the number of computers in a home increasing is 0.01153.
What is differentiation?
The rate of change of a function with respect to the variable is called differentiation. It can be increasing or decreasing.
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) Computer per household will be
Mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000
Then
q = 0.3458 ln (30000) − 3.045
q = 0.5187
b) The rate of increase in computers will be
[tex]\rm \dfrac{dq}{dt} = \dfrac{d}{dt} 0.3458lnx -3.045\\\\\\\dfrac{dq}{dt} = 0.3458 * \dfrac{1}{x} *\dfrac{dx}{dt}\\\\\\\dfrac{dq}{dt} = 0.3458 * \dfrac{1}{30000} *1000\\\\\\\dfrac{dq}{dt} = 0.01153[/tex]
More about the differentiation link is given below.
https://brainly.com/question/24062595
