Answer:
- 8 minutes
- 2 pounds
- 13 shifts
Step-by-step explanation:
These are problems in application of proportion. In each case, two related quantities are given, and you're asked to find the corresponding value when one of those quantities is changed. You do this by assuming the related quantities always have the same ratio. You need to be conversant in arithmetic using fractions.
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1)
The ratio of time to miles is presumed to be constant.
time/distance = (28 minutes)/(3.5 miles) = (m minutes)/(1 mile)
Multiply by the unwanted denominator. Same units cancel.
[tex]m\text{ min}=\dfrac{(28\text{ min})(1\text{ mi})}{(3.5\text{ mi})}=\dfrac{28}{3.5}\text{ min}=\boxed{8\text{ min}}[/tex]
It takes Sean 8 minutes to run 1 mile.
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2)
The ratio of flour to recipes is presumed to be constant.
flour/recipes = (8.25 lb)/(4.125 recipes) = q/(1 recipe)
As in the previous problem, we multiply by the unwanted denominator to get ...
q = (8.25 lb)(1 recipe)/(4.125 recipes) = (8.25/4.125) lb = 2 lb
2 lb of flour is used to make 1 recipe.
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3)
You can think of this as a units conversion problem. We have a quantity of hours per day, and we want a quantity of shifts per day. To get that, we multiply by a fraction with shifts in the numerator (the unit we want) and hours in the denominator (the unit we want to cancel).
The value of this multiplier fraction must be 1, which means its numerator and its denominator must represent the same quantity (but in different units).
We know that 1 shift = 3/4 hour, so our conversion fraction can be ...
(1 shift)/(0.75 hour)
The converted value is ...
[tex]\dfrac{9.75\text{ h}}{1\text{ day}}\times\dfrac{1\text{ shift}}{0.75\text{ h}}=\dfrac{9.75\text{ h}}{0.75\text{ h}}\,\dfrac{\text{shift}}{\text{day}}=\boxed{13\text{ shift/day}}[/tex]
Note that units of hours (h) cancel.
