Answer:
The water level dropped [tex]\frac{1}{12}[/tex] inch each hour
Step-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 hour
The water level dropped [tex]\frac{1}{12}[/tex] inch each hour
Answer:
[tex]\frac{1}{12}[/tex] inches per hour.
Step-by-step explanation:
Between 10 P.M. and 7:45 A.M., the water level in a swimming pool decreased by [tex]\frac{13}{16}[/tex] inch.
First we calculate the hours between 10:00 P.M and 7:45 A.M.
10:00 P.M. = 22:00 P.M.
10:00 p.m. + 2 hours = 12:00 a.m.
so 2 hours + 7 hours and 45 minutes = 9 hours 45 minutes
9 hours 45 minutes = [tex]9\frac{3}{4}[/tex] = [tex]\frac{39}{4}[/tex] hours
It takes [tex]\frac{39}{4}[/tex] hours to decrease the water level by [tex]\frac{13}{16}[/tex] inches.
In 1 hour the water level would decrease = [tex]\frac{\frac{13}{16}}{\frac{39}{4} }[/tex]
= [tex]\frac{13}{16}[/tex] × [tex]\frac{4}{39}[/tex]
= [tex]\frac{13}{4\times 39}[/tex]
= [tex]\frac{13}{156}[/tex]
= [tex]\frac{1}{12}[/tex] inches/hour
The water level decreased at a [tex]\frac{1}{12}[/tex] inches each hour.
