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e-lub [12.9K]
3 years ago
15

PLEASE ANSWER THIS IS URGENT!

Mathematics
1 answer:
lawyer [7]3 years ago
7 0

Answer:

question 12

A

question 13

c

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In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th
WITCHER [35]

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, \frac{dx}{dt} = $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 \frac{dq}{dt}

\frac{dq}{dt} = \frac{d(0.3458 ln x - 3.045)}{dt}

or

\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0

on substituting the values, we get

\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000

or

= 0.01153 computers

6 0
3 years ago
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