At 20 m from a localized sound source you measure the intensity level as 75 db. how far away must you be for theperceived loud-ness to drop in half [i. e., to an intensity level of 65 db)?

21.48m

Explanation:

Using the inverse square law, the intensity of sound is inversely proportional to the square of the distance.

Intensity ∝ 1/distance²

let l represent intensity and d represent distanc

l₁/l₂ = d₂²/d₁² 1

given;

l₁ = 75dB

l₂ = 65dB

d₁ = 20 m

substituting into eqn 1, we have;

75/65 = d₂²/20²

d₂² = 75 × 20²/65

d₂² = 461.54

d₂ = √461.54

d₂ = 21.48m

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form