Answer:
p + 0.0001q = 5
Step-by-step explanation:
Data provided in the question:
When Price, p₁ = $4.77 ; Demand, q₁ = 2,300
and,
When Price, p₂ = 3,300 ; Demand, q₂ = 3,300
considering the above as the points on the line,
now, the general equation for a line is given as:
![\frac{q_1-q}{p_1-p}=\frac{q_2-q_1}{p_2-p_1}](https://tex.z-dn.net/?f=%5Cfrac%7Bq_1-q%7D%7Bp_1-p%7D%3D%5Cfrac%7Bq_2-q_1%7D%7Bp_2-p_1%7D)
on substituting the respective values, we get
![\frac{2,300-q}{4.77-p}=\frac{3,300-2,300}{4.67-4.77}](https://tex.z-dn.net/?f=%5Cfrac%7B2%2C300-q%7D%7B4.77-p%7D%3D%5Cfrac%7B3%2C300-2%2C300%7D%7B4.67-4.77%7D)
or
![\frac{2,300-q}{4.77-p}=\frac{1000}{-0.1}](https://tex.z-dn.net/?f=%5Cfrac%7B2%2C300-q%7D%7B4.77-p%7D%3D%5Cfrac%7B1000%7D%7B-0.1%7D)
or
2,300 - q = -10,000(4.77 - p)
or
⇒ 2,300 - q = -47700 + 10,000p
or
⇒ 10,000p + q = 2300 + 47700
or
⇒ 10,000p + q = 50000
or
⇒ p + 0.0001q = 5