MATH SOLVE

4 months ago

Q:
# Between 10 P.M. and7:45 A.M., the water level in a swimming pooldecreased by 13/16 inch.Assuming that the water level decreased at aconstant rate, how much did it drop each hour?

Accepted Solution

A:

Answer:The water level dropped [tex]\frac{1}{12}[/tex] inch each hourStep-by-step explanation:- Between 10 P.M. and 7:45 A.M., the water level in a swimming pool decreased by 13/16 inch- Assuming that the water level decreased at a constant rate- We need to find the drop each hour that means the unit rate of decreased of the level of the water- At first lets hind how many hours between 10 P.M. and 7:45 A.M.∵ Between 10 P.M. and mid-night 2 hours∵ Between mid-night and 7:45 A.M. 7 hours and 45 minutes- Lets change 7 hours and 45 minutes to hours∵ 1 hour = 60 minutes∴ 45 minutes = 45 ÷ 60 = [tex]\frac{3}{4}[/tex] hours∴ 7 hours and 45 minutes = [tex]7\frac{3}{4}[/tex] hours∴ The total hours between 10 P.M. and 7:45 A.M. = 2 + [tex]7\frac{3}{4}[/tex] hours∴ The total hours between 10 P.M. and 7:45 A.M. = [tex]9\frac{3}{4}[/tex]∵ The unit rate of decreased = The decreased level ÷ total hours∵ The decreased level is [tex]\frac{13}{16}[/tex] inche∵ The total hours = [tex]9\frac{3}{4}[/tex] hours- Lets change the mixed number [tex]9\frac{3}{4}[/tex] to improper fraction∵ [tex]9\frac{3}{4}[/tex] = [tex]\frac{(9*4)+3}{4}[/tex]∴ [tex]9\frac{3}{4}[/tex] = [tex]\frac{39}{4}[/tex]∵ The unit rate of decreased = [tex]\frac{13}{16}[/tex] ÷ [tex]\frac{39}{4}[/tex]- To solve the division of 2 fractions change the division sign to multiplication sign and reciprocal the fraction after the division sign∴ The unit rate of decreases = [tex]\frac{13}{16}[/tex] × [tex]\frac{4}{39}[/tex] ∴ The unit rate of decreases = [tex]\frac{1}{12}[/tex] inch per hourThe water level dropped [tex]\frac{1}{12}[/tex] inch each hour