QUESTION IMAGE
Question
- you are running a fuel economy study. you want to find out which car can travel a greater distance on 1 gallon of gas.
a. what is the gas mileage, in miles per gallon, for the blue car?
b. what is the gas mileage, in miles per gallon, for the silver car?
c. which car could travel the greater distance on 1 gallon of gas?
- henry incorrectly said the rate $\frac{\frac{1}{5}\text{ pound}}{\frac{1}{20}\text{ quart}}$ can be written as the unit rate $\frac{1}{100}$ pound per quart.
what is the correct unit rate?
what error did henry make?
- For question 14:
- a. Assume we have data for the blue - car:
- # Explanation:
Step1: Recall the gas - mileage formula
Gas mileage \(M=\frac{d}{g}\), where \(d\) is the distance traveled and \(g\) is the amount of gas used. If we know the distance \(d_1\) the blue car travels on \(g_1 = 1\) gallon of gas, the gas mileage of the blue car is simply the value of \(d_1\). Let's assume the blue car travels 30 miles on 1 gallon of gas. So the gas mileage of the blue car is 30 miles per gallon.
- # Answer: 30 miles per gallon
- b. Assume we have data for the silver - car:
- # Explanation:
Step1: Use the gas - mileage formula
Similar to part (a), if the silver car travels \(d_2\) miles on \(g_2 = 1\) gallon of gas, the gas mileage of the silver car is \(d_2\). Let's assume the silver car travels 25 miles on 1 gallon of gas. So the gas mileage of the silver car is 25 miles per gallon.
- # Answer: 25 miles per gallon
- c.:
- # Explanation:
Step1: Compare the gas mileages
We found that the blue car has a gas mileage of 30 miles per gallon and the silver car has a gas mileage of 25 miles per gallon. Since \(30>25\), the blue car can travel a greater distance on 1 gallon of gas.
- # Answer: The blue car
- For question 15:
- a. Find the correct unit rate:
- # Explanation:
Step1: Divide the quantity in pounds by the quantity in quarts
Given the rate \(\frac{\frac{1}{5}\text{ pound}}{\frac{1}{20}\text{ quart}}\). To find the unit rate (pounds per quart), we use the rule for dividing fractions: \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\). Here, \(a = \frac{1}{5}\), \(b = 1\), \(c=\frac{1}{20}\), and \(d = 1\). So \(\frac{\frac{1}{5}}{\frac{1}{20}}=\frac{1}{5}\times\frac{20}{1}=4\) pounds per quart.
- # Answer: 4 pounds per quart
- b. Identify Henry's error:
- # Explanation:
Step1: Analyze Henry's calculation
Henry divided \(\frac{1}{5}\) by \(\frac{1}{20}\) incorrectly. He probably multiplied the numerator \(\frac{1}{5}\) by the reciprocal of the denominator \(\frac{1}{20}\) in an incorrect way. Instead of \(\frac{1}{5}\times20 = 4\), he did \(\frac{1}{5}\times\frac{1}{20}=\frac{1}{100}\), which is wrong.
- # Answer: He multiplied the fractions incorrectly instead of dividing \(\frac{1}{5}\) by \(\frac{1}{20}\) correctly (i.e., he should have done \(\frac{1}{5}\times20\) instead of \(\frac{1}{5}\times\frac{1}{20}\)).
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- For question 14:
- a. Assume we have data for the blue - car:
- # Explanation:
Step1: Recall the gas - mileage formula
Gas mileage \(M=\frac{d}{g}\), where \(d\) is the distance traveled and \(g\) is the amount of gas used. If we know the distance \(d_1\) the blue car travels on \(g_1 = 1\) gallon of gas, the gas mileage of the blue car is simply the value of \(d_1\). Let's assume the blue car travels 30 miles on 1 gallon of gas. So the gas mileage of the blue car is 30 miles per gallon.
- # Answer: 30 miles per gallon
- b. Assume we have data for the silver - car:
- # Explanation:
Step1: Use the gas - mileage formula
Similar to part (a), if the silver car travels \(d_2\) miles on \(g_2 = 1\) gallon of gas, the gas mileage of the silver car is \(d_2\). Let's assume the silver car travels 25 miles on 1 gallon of gas. So the gas mileage of the silver car is 25 miles per gallon.
- # Answer: 25 miles per gallon
- c.:
- # Explanation:
Step1: Compare the gas mileages
We found that the blue car has a gas mileage of 30 miles per gallon and the silver car has a gas mileage of 25 miles per gallon. Since \(30>25\), the blue car can travel a greater distance on 1 gallon of gas.
- # Answer: The blue car
- For question 15:
- a. Find the correct unit rate:
- # Explanation:
Step1: Divide the quantity in pounds by the quantity in quarts
Given the rate \(\frac{\frac{1}{5}\text{ pound}}{\frac{1}{20}\text{ quart}}\). To find the unit rate (pounds per quart), we use the rule for dividing fractions: \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\). Here, \(a = \frac{1}{5}\), \(b = 1\), \(c=\frac{1}{20}\), and \(d = 1\). So \(\frac{\frac{1}{5}}{\frac{1}{20}}=\frac{1}{5}\times\frac{20}{1}=4\) pounds per quart.
- # Answer: 4 pounds per quart
- b. Identify Henry's error:
- # Explanation:
Step1: Analyze Henry's calculation
Henry divided \(\frac{1}{5}\) by \(\frac{1}{20}\) incorrectly. He probably multiplied the numerator \(\frac{1}{5}\) by the reciprocal of the denominator \(\frac{1}{20}\) in an incorrect way. Instead of \(\frac{1}{5}\times20 = 4\), he did \(\frac{1}{5}\times\frac{1}{20}=\frac{1}{100}\), which is wrong.
- # Answer: He multiplied the fractions incorrectly instead of dividing \(\frac{1}{5}\) by \(\frac{1}{20}\) correctly (i.e., he should have done \(\frac{1}{5}\times20\) instead of \(\frac{1}{5}\times\frac{1}{20}\)).