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Answer :
To determine which function accurately represents the growth of water hyacinth in the testing pool, let's break down the problem step by step.
1. Understanding the Problem:
- We have an initial sample of 150 grams of water hyacinth introduced into a testing pool.
- The water hyacinth grows at a daily rate of 7%.
2. Calculating the Weekly Growth:
- Since there are 7 days in a week and growth is compounded daily at 7%, we want the function that models this weekly growth.
- The daily growth rate of 7% corresponds to multiplying the amount by [tex]\(1.07\)[/tex] each day.
3. Function Analysis:
- Now, let's examine each function option given:
- (A) [tex]\( f(t) = 150\left(1 + 0.07^{(1/7)}\right)^t \)[/tex]: This applies a daily rate incorrectly for weekly calculation.
- (B) [tex]\( g(t) = 150\left(1.07^{(1/7)}\right)^t \)[/tex]: Similar issue as with option A.
- (C) [tex]\( h(t) = 150\left(1 + 0.07^{(7)}\right)^t \)[/tex]: The growth rate here doesn’t follow the correct compounding process for a week.
- (D) [tex]\( k(t) = 150\left(1.07^{(7)}\right)^t \)[/tex]: This function calculates the growth over 7 days, compounding the growth correctly for an entire week.
4. Evaluating Each Option:
- We need to determine the amount present after one week (t = 1).
- The correct function should give us the real growth at the end of one week. Comparing the results:
- (A) and (B) provide incorrect interpretations for weekly growth, as they apply daily transformations improperly.
- (C) does not compute the weekly compounded growth correctly at all.
- (D) gives a plausible result—by correctly compounding the daily growth rate over 7 days and multiplying that factor for each week.
The function [tex]\( k(t) = 150\left(1.07^{(7)}\right)^t \)[/tex], option (D), best fits the conditions of compounding a daily growth over a week's period.
Therefore, the correct function that represents the amount of water hyacinth [tex]\(t\)[/tex] weeks after introduction is represented by option D.
1. Understanding the Problem:
- We have an initial sample of 150 grams of water hyacinth introduced into a testing pool.
- The water hyacinth grows at a daily rate of 7%.
2. Calculating the Weekly Growth:
- Since there are 7 days in a week and growth is compounded daily at 7%, we want the function that models this weekly growth.
- The daily growth rate of 7% corresponds to multiplying the amount by [tex]\(1.07\)[/tex] each day.
3. Function Analysis:
- Now, let's examine each function option given:
- (A) [tex]\( f(t) = 150\left(1 + 0.07^{(1/7)}\right)^t \)[/tex]: This applies a daily rate incorrectly for weekly calculation.
- (B) [tex]\( g(t) = 150\left(1.07^{(1/7)}\right)^t \)[/tex]: Similar issue as with option A.
- (C) [tex]\( h(t) = 150\left(1 + 0.07^{(7)}\right)^t \)[/tex]: The growth rate here doesn’t follow the correct compounding process for a week.
- (D) [tex]\( k(t) = 150\left(1.07^{(7)}\right)^t \)[/tex]: This function calculates the growth over 7 days, compounding the growth correctly for an entire week.
4. Evaluating Each Option:
- We need to determine the amount present after one week (t = 1).
- The correct function should give us the real growth at the end of one week. Comparing the results:
- (A) and (B) provide incorrect interpretations for weekly growth, as they apply daily transformations improperly.
- (C) does not compute the weekly compounded growth correctly at all.
- (D) gives a plausible result—by correctly compounding the daily growth rate over 7 days and multiplying that factor for each week.
The function [tex]\( k(t) = 150\left(1.07^{(7)}\right)^t \)[/tex], option (D), best fits the conditions of compounding a daily growth over a week's period.
Therefore, the correct function that represents the amount of water hyacinth [tex]\(t\)[/tex] weeks after introduction is represented by option D.
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