chlorophyll in plants
2. the trophic status of a body of water uses the amount of nutrients in the water to predict the amount of plant growth. your science class measures the amount of chlorophyll in a rectangular lake to determine its trophic status. the lake is 80 yards wide and has an area of (80x + 5600) square yards.
a. write and simplify an expression that represents the perimeter (in yards) of the lake.
b. the value of x is 90. what is the length of the lake?
c. your teacher wants to collect 5 samples of water at the edge of the lake. each sample should be taken at least 20 yards apart. is it possible to take all of the samples along the length of the lake? explain.
3. the table shows the amount of chlorophyll for each trophic class.
| chlorophyll level (parts per billion) | trophic class | biological productivity |
| ---- | ---- | ---- |
| 0–2.6 | oligotrophic | low |
| 2.6–20 | mesotrophic | intermediate |
| 20–56 | eutrophic | high |
| 56–155 | hypereutrophic | severe |
a. the chlorophyll levels from the samples taken are 5.7, 12.3, 17.1, 20.1, and 23.1 parts per billion. what trophic class is indicated by the mean chlorophyll level?
b. give an example of 5 sample chlorophyll levels such that the mean chlorophyll level would classify the lake as eutrophic.

Question

chlorophyll in plants
2. the trophic status of a body of water uses the amount of nutrients in the water to predict the amount of plant growth. your science class measures the amount of chlorophyll in a rectangular lake to determine its trophic status. the lake is 80 yards wide and has an area of (80x + 5600) square yards.
 a. write and simplify an expression that represents the perimeter (in yards) of the lake.
 b. the value of x is 90. what is the length of the lake?
 c. your teacher wants to collect 5 samples of water at the edge of the lake. each sample should be taken at least 20 yards apart. is it possible to take all of the samples along the length of the lake? explain.
3. the table shows the amount of chlorophyll for each trophic class.
| chlorophyll level (parts per billion) | trophic class | biological productivity |
| ---- | ---- | ---- |
| 0–2.6 | oligotrophic | low |
| 2.6–20 | mesotrophic | intermediate |
| 20–56 | eutrophic | high |
| 56–155 | hypereutrophic | severe |
 a. the chlorophyll levels from the samples taken are 5.7, 12.3, 17.1, 20.1, and 23.1 parts per billion. what trophic class is indicated by the mean chlorophyll level?
 b. give an example of 5 sample chlorophyll levels such that the mean chlorophyll level would classify the lake as eutrophic.

Accepted Answer

### Part 2a
# Explanation:
## Step1: Recall rectangle area formula
The area of a rectangle is $ A = \text{width} \times \text{length} $. Given width $ w = 80 $ yards and area $ A = 80x + 5600 $, we find length $ l $ by solving $ 80l = 80x + 5600 $. Divide both sides by 80: $ l=\frac{80x + 5600}{80}=x + 70 $.
## Step2: Recall rectangle perimeter formula
Perimeter of a rectangle is $ P = 2(w + l) $. Substitute $ w = 80 $ and $ l = x + 70 $: $ P = 2(80+(x + 70)) $.
## Step3: Simplify the perimeter expression
Simplify inside the parentheses: $ 80+(x + 70)=x + 150 $. Then multiply by 2: $ P = 2(x + 150)=2x + 300 $.
# Answer: $ 2x + 300 $

### Part 2b
# Explanation:
## Step1: Use the length formula from 2a
From 2a, length $ l = x + 70 $.
## Step2: Substitute $ x = 90 $
Substitute $ x = 90 $ into $ l = x + 70 $: $ l = 90 + 70 = 160 $.
# Answer: 160 yards

### Part 2c
# Explanation:
## Step1: Determine the required spacing
For 5 samples, there are $ 5 - 1 = 4 $ intervals. Each interval needs to be at least 20 yards. So total required length for spacing is $ 4\times20 = 80 $ yards.
## Step2: Compare with lake length
From 2b, the length of the lake is 160 yards. Since $ 80\leq160 $, there is enough space.
# Answer: Yes, because the total required spacing (80 yards) is less than the lake's length (160 yards).

### Part 3a
# Explanation:
## Step1: Calculate the mean chlorophyll level
Mean $ \bar{x}=\frac{5.7 + 12.3 + 17.1 + 20.1 + 23.1}{5} $. Sum the values: $ 5.7+12.3 = 18 $, $ 18+17.1 = 35.1 $, $ 35.1+20.1 = 55.2 $, $ 55.2+23.1 = 78.3 $. Then divide by 5: $ \bar{x}=\frac{78.3}{5}=15.66 $.
## Step2: Classify the mean using the table
The mean 15.66 is in the range $ 2.6 - 20 $, which corresponds to Mesotrophic.
# Answer: Mesotrophic

### Part 3b
# Explanation:
## Step1: Recall the Eutrophic range
Eutrophic range is $ 20 - 56 $ parts per billion. The mean of 5 samples must be in this range. Let's choose 5 values whose mean is in $ 20 - 56 $. For example, let the samples be 20, 20, 30, 40, 40.
## Step2: Calculate the mean of the example
Mean $ \bar{x}=\frac{20 + 20 + 30 + 40 + 40}{5}=\frac{150}{5}=30 $, which is in $ 20 - 56 $.
# Answer: Example: 20, 20, 30, 40, 40 (any 5 values with a mean between 20 and 56 will work)

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QUESTION IMAGE

Question

Explanation:

Part 2a

Step1: Recall rectangle area formula

Step2: Recall rectangle perimeter formula

Step3: Simplify the perimeter expression

Step1: Use the length formula from 2a

Step2: Substitute \( x = 90 \)

Step1: Determine the required spacing

Step2: Compare with lake length

Answer:

Part 2b

chlorophyll level (parts per billion)	trophic class	biological productivity
2.6–20	mesotrophic	intermediate
20–56	eutrophic	high
56–155	hypereutrophic	severe