This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio ($1 \text{ mi}=5280 \text{ ft}$ gives $5280 \text{ ft}/1 \text{ mi}$, $1 \text{ hr}=3600 \text{ s}$ gives $1 \text{ hr}/3600 \text{ s}$), multiply: $60 \text{ mi/hr} \times(5280 \text{ ft}/1 \text{ mi}) \times(1 \text{ hr}/3600 \text{ s})=88 \text{ ft/s}$ (mi and hr cancel leaving ft/s). Direction: for compound units like speed, chain conversions to cancel step-by-step. Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: $60 \text{ mph}$ to ft/sec uses $60 \text{ mi/hr} \times(5280 \text{ ft}/1 \text{ mi}) \times(1 \text{ hr}/3600 \text{ sec})$, mi cancels, hr cancels, result: $(60\times5280/3600) \text{ ft/sec}=88 \text{ ft/sec}$; or $3 \text{ feet}$ to inches: $3 \text{ ft} \times(12 \text{ in}/1 \text{ ft})=36 \text{ inches}$; or area $5 \text{ ft} \times 3 \text{ ft}=15 \text{ ft}^2$ (units multiply: ft×ft=ft² square feet). The correct conversion is $60 \text{ mi/hr} \times(5280 \text{ ft} / 1 \text{ mi}) \times(1 \text{ hr} / 3600 \text{ s}) = 88 \text{ ft/s}$, with miles and hours canceling out. A common error is wrong direction (dividing when should multiply for certain factors), conversion factor wrong (using $5000 \text{ ft/mi}$), units not canceled (leaving mi/ft or similar), arithmetic error ($60\times5280/3600=$ wrong calc), or omitting a factor (forgetting time conversion). Process: (1) identify units (start: mi/hr, target: ft/s), (2) find conversions ($1 \text{ mi}=5280 \text{ ft}$, $1 \text{ hr}=3600 \text{ s}$), (3) set up with cancellation ($60 \text{ mi/hr} \times(5280 \text{ ft}/1 \text{ mi}) \times(1 \text{ hr}/3600 \text{ s}$)), (4) calculate ($60\times5280/3600=88 \text{ ft/s}$), (5) verify units (answer should be in ft/s✓). Dimensional analysis: write conversion factors as fractions with units, multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio ($1 \text{ hr}=60 \text{ min}$ gives ratio $60 \text{ min}$ per $1 \text{ hr}$), multiply: $1.5 \text{ hr} \times(60 \text{ min}/1 \text{ hr})=90 \text{ min}$ (hr cancels: hr in numerator and denominator divide out leaving min). Direction: larger unit→smaller unit multiply (hr→min: ×60), smaller→larger divide (min→hr: ÷60, or ×(1/60)). Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert $1.5$ hours to minutes using $1 \text{ hr}=60 \text{ min}$, multiply: $1.5 \text{ hr} \times(60 \text{ min} / 1 \text{ hr})=1.5\times60 \text{ min}=90$ minutes (hours cancel); or $3$ feet to inches: $3 \text{ ft} \times(12 \text{ in}/1 \text{ ft})=36$ inches; or area $5 \text{ ft} \times 3 \text{ ft}=15 \text{ ft}^2$ (units multiply: ft×ft=ft² square feet). The correct conversion is $1.5 \text{ hr} \times(60 \text{ min} / 1 \text{ hr}) = 90 \text{ min}$, with hour units canceling out. A common error is wrong direction (dividing when should multiply: $1.5\div60=0.025 \text{ hr}$ instead of min), conversion factor wrong (using $100 \text{ min}/\text{hr}$), units not canceled (answer $90 \text{ hr}\cdot\text{min}$ instead of $90 \text{ min}$), arithmetic error ($1.5\times60=80$), or adds instead of multiplies ($1.5+60=61.5$ nonsense). Process: (1) identify units (start: hr, target: min), (2) find conversion ($1 \text{ hr}=60 \text{ min}$), (3) set up with cancellation ($1.5 \text{ hr}\times(60 \text{ min}/1 \text{ hr}$, hr cancels), (4) calculate ($1.5\times60=90 \text{ min}$), (5) verify units (answer should be in minutes✓). Dimensional analysis: write conversion factors as fractions with units ($60 \text{ min}/1 \text{ hr}$), multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 min=60 s gives ratio 60 s per 1 min), multiply: 2.25 min×(60 s/1 min)=135 s (min cancels: min in numerator and denominator divide out leaving s). Direction: larger unit→smaller unit multiply (min→s: ×60), smaller→larger divide (s→min: ÷60, or ×(1/60)). Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert 2.25 minutes to seconds using 1 min=60 s, multiply: 2.25 min×(60 s/1 min)=2.25×60 s=135 seconds (minutes cancel); or 3 feet to inches: 3 ft×(12 in/1 ft)=36 inches; or area 5 ft×3 ft=15 ft² (units multiply: ft×ft=ft² square feet). The correct conversion is 2.25 min × (60 s / 1 min) = 135 s, with minute units canceling out. A common error is wrong direction (dividing when should multiply: 2.25÷60=0.0375 min instead of s), conversion factor wrong (using 100 s/min), units not canceled (answer 135 min·s instead of 135 s), arithmetic error (2.25×60=120), or adds instead of multiplies (2.25+120=122.25 nonsense). Process: (1) identify units (start: min, target: s), (2) find conversion (1 min=60 s), (3) set up with cancellation (2.25 min×(60 s/1 min), min cancels), (4) calculate (2.25×60=135 s), (5) verify units (answer should be in seconds✓). Dimensional analysis: write conversion factors as fractions with units (60 s/1 min), multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 hr=60 min gives ratio 60 min per 1 hr), multiply: 2.5 hr×(60 min/1 hr)=150 min (hr cancels: hr in numerator and denominator divide out leaving min). Direction: larger unit→smaller unit multiply (hr→min: ×60), smaller→larger divide (min→hr: ÷60, or ×(1/60)). For example, convert 2.5 hours to minutes using 1 hr=60 min, multiply: 2.5 hr×(60 min/1 hr)=2.5×60 min=150 minutes (hours cancel). The correct conversion is 2.5 hr × (60 min/1 hr) = 150 min, with hour units canceling out. A common error is reversing the direction, like dividing 2.5 by 60 to get about 0.0417 hr instead of multiplying, or ignoring the decimal and using only 2 hr to get 120 min, or adding instead of multiplying like 2.5+60=62.5. The process is: (1) identify units (start: hr, target: min), (2) find conversion (1 hr=60 min), (3) set up with cancellation (2.5 hr×(60 min/1 hr), hr cancels), (4) calculate (2.5×60=150 min), (5) verify units (answer in minutes✓). Dimensional analysis ensures units cancel properly, and common mistakes include arithmetic errors or not handling decimal values correctly.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 lb=16 oz gives ratio 16 oz per 1 lb), multiply: 3 lb×(16 oz/1 lb)=48 oz (lb cancels: lb in numerator and denominator divide out leaving oz). Direction: larger unit→smaller unit multiply (lb→oz: ×16), smaller→larger divide (oz→lb: ÷16, or ×(1/16)). Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert 3 pounds to ounces using 1 lb=16 oz, multiply: 3 lb×(16 oz/1 lb)=3×16 oz=48 ounces (pounds cancel); or 2.5 hours to minutes: 2.5 hr×(60 min/1 hr)=150 minutes; or area 5 ft×3 ft=15 ft² (units multiply: ft×ft=ft² square feet). The correct conversion is 3 lb × (16 oz / 1 lb) = 48 oz, with pound units canceling out. A common error is wrong direction (dividing when should multiply: 3÷16=0.1875 lb instead of oz), conversion factor wrong (using 10 oz/lb), units not canceled (answer 48 lb·oz instead of 48 oz), arithmetic error (3×16=42), or adds instead of multiplies (3+16=19 nonsense). Process: (1) identify units (start: lb, target: oz), (2) find conversion (1 lb=16 oz), (3) set up with cancellation (3 lb×(16 oz/1 lb), lb cancels), (4) calculate (3×16=48 oz), (5) verify units (answer should be in ounces✓). Dimensional analysis: write conversion factors as fractions with units (16 oz/1 lb), multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 mi=5280 ft gives ratio 5280 ft per 1 mi), multiply: 1.5 mi×(5280 ft/1 mi)=7920 ft (mi cancels: mi in numerator and denominator divide out leaving ft). Direction: larger unit→smaller unit multiply (mi→ft: ×5280). For example, convert 1.5 miles to feet using 1 mi=5280 ft, multiply: 1.5 mi×(5280 ft/1 mi)=1.5×5280 ft=7920 feet (miles cancel). The correct conversion is 1.5 mi × (5280 ft/1 mi) = 7920 ft, with mile units canceling out. A common error is reversing the direction, like dividing 1.5 by 5280 to get about 0.000284 mi, or using approximations like 1.5×6000=9000 then subtracting arbitrarily, or subtracting like 5280-1760=3520. The process is: (1) identify units (start: mi, target: ft), (2) find conversion (1 mi=5280 ft), (3) set up with cancellation (1.5 mi×(5280 ft/1 mi), mi cancels), (4) calculate (1.5×5280=7920 ft), (5) verify units (answer in feet✓). Common conversions to memorize include 1 mi=5280 ft, and mistakes include arithmetic errors with decimals.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 m=100 cm gives ratio 1 m per 100 cm), multiply: 450 cm×(1 m/100 cm)=4.5 m (cm cancels: cm in numerator and denominator divide out leaving m). Direction: smaller unit→larger unit divide (cm→m: ÷100, or ×(1/100)), larger→smaller multiply (m→cm: ×100). Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert 450 centimeters to meters using 1 m=100 cm, multiply: 450 cm×(1 m/100 cm)=450/100 m=4.5 meters (centimeters cancel); or 2.5 hours to minutes: 2.5 hr×(60 min/1 hr)=150 minutes; or area 5 ft×3 ft=15 ft² (units multiply: ft×ft=ft² square feet). The correct conversion is 450 cm × (1 m / 100 cm) = 4.5 m, with centimeter units canceling out. A common error is wrong direction (multiplying when should divide: 450×100=45,000 cm instead of m), conversion factor wrong (using 1 m=10 cm), units not canceled (answer 4.5 cm·m instead of 4.5 m), arithmetic error (450/100=4), or adds instead of multiplies (450+100=550 nonsense). Process: (1) identify units (start: cm, target: m), (2) find conversion (1 m=100 cm), (3) set up with cancellation (450 cm×(1 m/100 cm), cm cancels), (4) calculate (450/100=4.5 m), (5) verify units (answer should be in meters✓). Dimensional analysis: write conversion factors as fractions with units (1 m/100 cm), multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting from milliliters to liters: use the conversion factor as a ratio ($1 \text{ L} = 1000 \text{ mL}$ gives $1 \text{ L}/1000 \text{ mL}$), multiply: $750 \text{ mL} \times(1 \text{ L}/1000 \text{ mL}) = 0.75 \text{ L}$ ($\text{mL}$ cancels leaving $\text{L}$). Direction: smaller unit→larger unit divide ($\text{mL}$→$\text{L}$: ÷1000, or ×(1/1000)). Units multiply/divide: length×length=area ($\text{ft} \times \text{ft} = \text{ft}^2$), distance÷time=speed ($\text{mi} \div \text{hr} = \text{mi/hr}$ or mph), units treated algebraically. Example: convert 3 feet to inches using $1 \text{ ft} = 12 \text{ in}$, multiply: $3 \text{ ft} \times(12 \text{ in}/1 \text{ ft}) = 3 \times 12 \text{ in} = 36 \text{ inches}$ (feet cancel); or 2.5 hours to minutes: $2.5 \text{ hr} \times(60 \text{ min}/1 \text{ hr}) = 150 \text{ minutes}$; or area $5 \text{ ft} \times 3 \text{ ft} = 15 \text{ ft}^2$ (units multiply: $\text{ft} \times \text{ft} = \text{ft}^2$ square feet). The correct conversion is $750 \text{ mL} / 1000 = 0.75 \text{ L}$. A common error is multiplying instead of dividing, like $750 \times 1000 = 750,000$ (way too large), or misplaced decimal (7.5 instead of 0.75). Process: (1) identify units (start: mL, target: L), (2) find conversion ($1 \text{ L} = 1000 \text{ mL}$), (3) set up with cancellation ($750 \text{ mL} \times(1 \text{ L}/1000 \text{ mL}$, mL cancels), (4) calculate ($750/1000 = 0.75 \text{ L}$), (5) verify units (answer should be in liters✓). Dimensional analysis: write conversion factors as fractions with units ($1 \text{ L}/1000 \text{ mL}$), multiply so units cancel leaving desired unit.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting: use conversion factor as ratio (1 ft=12 in gives ratio 12 in per 1 ft), multiply for area: first 2 ft×(12 in/1 ft)=24 in, then 8 in×24 in=192 in² (units multiply: in×in=in²). Direction: larger unit→smaller unit multiply (ft→in: ×12), and remember area units square when multiplying lengths. Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert 2 feet to inches using 1 ft=12 in, multiply: 2 ft×(12 in/1 ft)=24 inches (feet cancel), then area 8 in×24 in=192 in² (units multiply: in×in=in² square inches); or 2.5 hours to minutes: 2.5 hr×(60 min/1 hr)=150 minutes; or area 5 ft×3 ft=15 ft². The correct conversion is 2 ft × (12 in / 1 ft) = 24 in, then 8 in × 24 in = 192 in², with feet canceling and units squaring properly. A common error is wrong direction (dividing height), conversion factor wrong, units not squared (answer 192 in instead of in²), arithmetic error (8×24=180), area units wrong (192 ft² not in²), or adds instead of multiplies. Process: (1) identify units (height: ft to in, then area in in²), (2) find conversion (1 ft=12 in), (3) set up conversion (2 ft×(12 in/1 ft)), (4) calculate area (8×24=192 in²), (5) verify units (answer in square inches✓). Dimensional analysis: write conversion factors as fractions with units, ensure multiplied units result in squared form for area.

This question tests converting measurement units using ratio reasoning (conversion factors as ratios), unit cancellation (dimensional analysis), and manipulating units in multiplication/division. Converting height from feet to inches: use the conversion factor as a ratio (1 ft=12 in gives 12 in/1 ft), multiply: 3 ft×(12 in/1 ft)=36 in (ft cancels leaving in), then area: 8 in × 36 in = 288 in² (units multiply to square inches). Direction: larger unit→smaller unit multiply (ft→in: ×12), and for area, ensure both dimensions in same unit. Units multiply/divide: length×length=area (ft×ft=ft²), distance÷time=speed (mi÷hr=mi/hr or mph), units treated algebraically. Example: convert 3 feet to inches using 1 ft=12 in, multiply: 3 ft×(12 in/1 ft)=3×12 in=36 inches (feet cancel); or 2.5 hours to minutes: 2.5 hr×(60 min/1 hr)=150 minutes; or area 5 ft×3 ft=15 ft² (units multiply: ft×ft=ft² square feet). The correct area is 8 in × 36 in = 288 in² after conversion. A common error is not converting units before multiplying, like 8 in × 3 ft = 24 in·ft (mixed units), or wrong arithmetic (8×36=96, too small). Process: (1) identify units (height: ft to in), (2) find conversion (1 ft=12 in), (3) convert (3 ft×12 in/ft=36 in), (4) calculate area (8×36=288 in²), (5) verify units (answer should be in in²✓). Multiplying lengths: ft×ft=ft² (area in square feet), but here convert to in².