Compare ways to protect biodiversity and choose the best
Grade 9 Science Worksheet
Circle the best answer.
Read this first
Use this
Decision matrix (model data), scores 1 to 5, a higher score is better in every column. Weights: benefit to species counts 3 times, low cost counts 2 times, few side effects counts 1 time. Protected area, benefit to species 5, low cost 2, few side effects 4; Wildlife corridor, benefit to species 4, low cost 3, few side effects 4; Seed bank, benefit to species 2, low cost 5, few side effects 5.
Pia weights the goals as shown. What is the weighted total for the protected area?
- 21 points
- 11 points
- 5 points
- 23 points: 15 + 4 + 4
Use this
Decision matrix (model data), scores 1 to 5, a higher score is better in every column. Weights: benefit to species counts 3 times, low cost counts 1 time, few side effects counts 1 time. Invasive-species control, benefit to species 3, low cost 4, few side effects 3; Wildlife corridor, benefit to species 4, low cost 3, few side effects 4; Reducing runoff, benefit to species 4, low cost 3, few side effects 5; Protected area, benefit to species 5, low cost 2, few side effects 4.
Ren gives benefit a weight of 3 and the other two columns a weight of 1. Which plan has the highest weighted total?
- Protected area: 21 points, the highest weighted total
- Reducing runoff: its plain sum, 12, is the highest of the rows
- Wildlife corridor: its weighted total is 19
- Invasive-species control: it has the top low-cost score, 4, so it is good value
Use this
Decision matrix (model data), scores 1 to 5, a higher score is better in every column. Weights: benefit to species counts 3 times, low cost counts 2 times, few side effects counts 1 time. Protected area, benefit to species 5, low cost 2, few side effects 4; Wildlife corridor, benefit to species 4, low cost 3, few side effects 4; Seed bank, benefit to species 2, low cost 5, few side effects 5.
Esme uses the weights shown. By how many points does the protected area beat the wildlife corridor?
- 0 points
- 2 points
- 3 points
- 1 point: 23 − 22
The wildlife corridor and three limits (model data)
measure corridor limit (at most) cost (million dollars) 7 8 land (hectares) 45 50 time to start (years) 6 4 Kira checks the wildlife corridor against each limit in the table. Which limit does it break?
- Time: 6 years is more than the 4-year limit on time
- Budget: 7 million dollars is too near the 8 million limit
- No limit: its cost of 7 million is under the budget of 8 million
- Land: 45 hectares is very close to the 50-hectare limit
Five plans and three limits (model data)
plan benefit score cost (million dollars) land (hectares) time to start (years) Protected area 5 9 120 5 Wildlife corridor 4 6 40 3 Invasive-species control 3 2 15 1 Seed bank 2 3 1 2 Reducing runoff 3 4 25 2 limit (at most) no limit 8 50 4 Milo reads the table and the three limits. How many of the five plans meet all three limits?
- 2 plans
- 4 plans
- 3 plans
- 5 plans
Beetles in patch B, a second model (model data)
generation beetles without a corridor beetles with a corridor 0 40 40 1 38 40 2 36 40 3 34 40 4 32 40 5 30 40 Tala runs a simulation of a beetle species in patch B. By generation 5, how many more beetles were there with a corridor?
- 10 more beetles: 40 − 30
- 40 more beetles
- 2 more beetles
- 8 more beetles
Patches in the cool band that beetles reached as it moved uphill, a range-shift model (model data)
decade patches reached without a corridor patches reached with a corridor 0 0 0 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 Bo reads a model in which the cool band on a slope moves uphill. Which statement does the output support?
- By decade 5 beetles reached 5 patches with a corridor and 1 without one
- Without a corridor beetles reached 5 patches by decade 5, and with one they reached 1
- The corridor slows the warming, so no other action is needed on the slope
- In decade 3 beetles reached the same number of patches with and without a corridor
Zee reads a model graph that tests removing an invasive plant. Which statement matches the graph?
- Cover with removal was 50% more than without, since 66 − 16 = 50
- Removal fixed the problem, so the plot needs no protection from any other driver
- In year 5, native cover was 66% with removal and 16% without, a gap of 50 points
- Native cover in the plot without removal stayed level at 40% for the five years
Use this
Decision matrix (model data), scores 1 to 5, a higher score is better in every column. Weights: benefit to species counts 3 times, low cost counts 2 times, few side effects counts 1 time. Protected area, benefit to species 5, low cost 2, few side effects 4; Wildlife corridor, benefit to species 4, low cost 3, few side effects 4; Reducing runoff, benefit to species 3, low cost 4, few side effects 5; Seed bank, benefit to species 2, low cost 5, few side effects 2.
Ari changes the weights in the matrix. Which statement is correct?
- The scores in the table must change too, because a heavier weight raises a score
- The scores stay the same, and the weighted totals can change
- A heavier weight on low cost makes the cheapest plan win
- The plan with the highest single score wins, whatever the weights
Four plans, weighted for a land limit of 50 hectares (model data)
plan benefit (×3) low cost (×2) few side effects (×1) land (hectares) Protected area 5 2 4 120 Wildlife corridor 4 4 3 40 Reducing runoff 3 4 5 25 Seed bank 1 5 5 1 Noah weights the columns as shown and has a land limit of 50 hectares. Which plan should be chosen?
- Wildlife corridor: 23 points, tied, and 40 hectares fits the limit
- Seed bank: it costs the least and needs just 1 hectare of land
- Protected area: it ties at 23 points and has the top benefit score, 5
- Reducing runoff: its plain sum, 12, is the highest in the table