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".
1 answer