Answer :
To determine the rate of change of Carol's distance traveled over time while cross-country skiing, we'll calculate the rate of change between each pair of consecutive data points. The rate of change is found using the formula:
[tex]\[ \text{Rate of Change} = \frac{\Delta \text{Distance}}{\Delta \text{Time}} \][/tex]
where [tex]\( \Delta \)[/tex] represents the change.
Step-by-Step Solution:
1. From 2 to 3 minutes:
- Distance at 2 minutes = [tex]\( \frac{1}{6} \)[/tex] miles
- Distance at 3 minutes = [tex]\( \frac{17}{48} \)[/tex] miles
- Change in distance: [tex]\( \frac{17}{48} - \frac{1}{6} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{17}{48} = \frac{17}{48} \][/tex]
[tex]\[ \frac{1}{6} = \frac{8}{48} \][/tex]
[tex]\[ \frac{17}{48} - \frac{8}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 3 - 2 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
2. From 3 to 4 minutes:
- Distance at 3 minutes = [tex]\( \frac{17}{48} \)[/tex] miles
- Distance at 4 minutes = [tex]\( \frac{13}{24} \)[/tex] miles
- Change in distance: [tex]\( \frac{13}{24} - \frac{17}{48} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{13}{24} = \frac{26}{48} \][/tex]
[tex]\[ \frac{26}{48} - \frac{17}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 4 - 3 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
3. From 4 to 5 minutes:
- Distance at 4 minutes = [tex]\( \frac{13}{24} \)[/tex] miles
- Distance at 5 minutes = [tex]\( \frac{35}{48} \)[/tex] miles
- Change in distance: [tex]\( \frac{35}{48} - \frac{13}{24} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{13}{24} = \frac{26}{48} \][/tex]
[tex]\[ \frac{35}{48} - \frac{26}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 5 - 4 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
4. From 5 to 6 minutes:
- Distance at 5 minutes = [tex]\( \frac{35}{48} \)[/tex] miles
- Distance at 6 minutes = [tex]\( \frac{11}{12} \)[/tex] miles
- Change in distance: [tex]\( \frac{11}{12} - \frac{35}{48} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{11}{12} = \frac{44}{48} \][/tex]
[tex]\[ \frac{44}{48} - \frac{35}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 6 - 5 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
The rate of change for each interval is approximately [tex]\( 0.1875 \)[/tex] miles per minute. Thus, Carol's rate of change in distance while skiing is constant at [tex]\( 0.1875 \)[/tex] miles per minute.
[tex]\[ \text{Rate of Change} = \frac{\Delta \text{Distance}}{\Delta \text{Time}} \][/tex]
where [tex]\( \Delta \)[/tex] represents the change.
Step-by-Step Solution:
1. From 2 to 3 minutes:
- Distance at 2 minutes = [tex]\( \frac{1}{6} \)[/tex] miles
- Distance at 3 minutes = [tex]\( \frac{17}{48} \)[/tex] miles
- Change in distance: [tex]\( \frac{17}{48} - \frac{1}{6} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{17}{48} = \frac{17}{48} \][/tex]
[tex]\[ \frac{1}{6} = \frac{8}{48} \][/tex]
[tex]\[ \frac{17}{48} - \frac{8}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 3 - 2 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
2. From 3 to 4 minutes:
- Distance at 3 minutes = [tex]\( \frac{17}{48} \)[/tex] miles
- Distance at 4 minutes = [tex]\( \frac{13}{24} \)[/tex] miles
- Change in distance: [tex]\( \frac{13}{24} - \frac{17}{48} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{13}{24} = \frac{26}{48} \][/tex]
[tex]\[ \frac{26}{48} - \frac{17}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 4 - 3 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
3. From 4 to 5 minutes:
- Distance at 4 minutes = [tex]\( \frac{13}{24} \)[/tex] miles
- Distance at 5 minutes = [tex]\( \frac{35}{48} \)[/tex] miles
- Change in distance: [tex]\( \frac{35}{48} - \frac{13}{24} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{13}{24} = \frac{26}{48} \][/tex]
[tex]\[ \frac{35}{48} - \frac{26}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 5 - 4 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
4. From 5 to 6 minutes:
- Distance at 5 minutes = [tex]\( \frac{35}{48} \)[/tex] miles
- Distance at 6 minutes = [tex]\( \frac{11}{12} \)[/tex] miles
- Change in distance: [tex]\( \frac{11}{12} - \frac{35}{48} \)[/tex]
- Simplify the subtraction:
[tex]\[ \frac{11}{12} = \frac{44}{48} \][/tex]
[tex]\[ \frac{44}{48} - \frac{35}{48} = \frac{9}{48} = \frac{3}{16} \][/tex]
- Change in time: 6 - 5 = 1 minute
- Rate of change: [tex]\( \frac{3/16}{1} = \frac{3}{16} \approx 0.1875 \)[/tex]
The rate of change for each interval is approximately [tex]\( 0.1875 \)[/tex] miles per minute. Thus, Carol's rate of change in distance while skiing is constant at [tex]\( 0.1875 \)[/tex] miles per minute.
