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The amount of fertilizer in a landscaping company's warehouse decreases at a rate of 2% per week. The amount of fertilizer in t...Asked by =
The amount of fertilizer in a landscaping company's warehouse decreases at a rate of 2 %2% per week. The amount of fertilizer in the warehouse was originally, 64 , 00064,000 cubic yards. Which function models the amount of fertilizer in cubic yards left after "w"w weeks?
"f" left parenthesis, "w" , right parenthesis equals 0.9 8 left parenthesis, 64000 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 0.9 8 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 1.0 2 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 1.9 8 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 0.9 8 left parenthesis, 64000 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 0.9 8 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 1.0 2 , right parenthesis Superscript, "w" , Baseline
"f" left parenthesis, "w" , right parenthesis equals 64 , 000 left parenthesis, 1.9 8 , right parenthesis Superscript, "w" , Baseline
Answers
Answered by
GPT-4o mini
To model the amount of fertilizer left after "w" weeks, we need to account for the reduction rate of 2% per week.
When something decreases by a percentage, we can model this with exponential decay. If the original amount is \( A_0 \) and it decreases by \( r \\%\) each period, the remaining amount after \( w \) periods can be described by the formula:
\[
A(w) = A_0 \times (1 - r)^w
\]
In this case, the original amount of fertilizer \( A_0 \) is 64,000 cubic yards and the rate of decrease \( r \) is 2%, or 0.02 when expressed as a decimal. Therefore, \( (1 - r) = 1 - 0.02 = 0.98 \).
Thus, the function that models the amount of fertilizer left after "w" weeks can be expressed as:
\[
f(w) = 64,000 \times (0.98)^w
\]
From the given options:
- \( f(w) = 0.98(64000)^w \) (doesn't model correctly, exponent applies to the wrong part)
- \( f(w) = 64,000 \times (0.98)^w \) (this is correct)
- \( f(w) = 64,000 \times (1.02)^w \) (this is incorrect, should be a decay, not growth)
- \( f(w) = 64,000 \times (1.98)^w \) (this is incorrect)
The correct function that models the amount of fertilizer in cubic yards left after \( w \) weeks is:
\[
f(w) = 64,000 \times (0.98)^w
\]
So the correct option is:
**"f(w) = 64,000 \times (0.98)^w"**.
When something decreases by a percentage, we can model this with exponential decay. If the original amount is \( A_0 \) and it decreases by \( r \\%\) each period, the remaining amount after \( w \) periods can be described by the formula:
\[
A(w) = A_0 \times (1 - r)^w
\]
In this case, the original amount of fertilizer \( A_0 \) is 64,000 cubic yards and the rate of decrease \( r \) is 2%, or 0.02 when expressed as a decimal. Therefore, \( (1 - r) = 1 - 0.02 = 0.98 \).
Thus, the function that models the amount of fertilizer left after "w" weeks can be expressed as:
\[
f(w) = 64,000 \times (0.98)^w
\]
From the given options:
- \( f(w) = 0.98(64000)^w \) (doesn't model correctly, exponent applies to the wrong part)
- \( f(w) = 64,000 \times (0.98)^w \) (this is correct)
- \( f(w) = 64,000 \times (1.02)^w \) (this is incorrect, should be a decay, not growth)
- \( f(w) = 64,000 \times (1.98)^w \) (this is incorrect)
The correct function that models the amount of fertilizer in cubic yards left after \( w \) weeks is:
\[
f(w) = 64,000 \times (0.98)^w
\]
So the correct option is:
**"f(w) = 64,000 \times (0.98)^w"**.
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