IDNLearn.com: Where your questions meet expert advice and community support. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
Let's examine both parts of the problem step-by-step:
### Part A: Analyzing the Price of Fuel A
The price \( f(x) \) of fuel A after \( x \) months is given by the function:
[tex]\[ f(x) = 2.15(0.98)^x \][/tex]
1. Determine if the price is increasing or decreasing:
- The function \( f(x) = 2.15(0.98)^x \) is in the form of an exponential decay function because the base of the exponent (0.98) is less than 1.
- This indicates that the price of fuel A is decreasing over time.
2. Find the rate of percentage decrease per month:
- In exponential decay functions of the form \( f(x) = a \cdot b^x \), where \( 0 < b < 1 \), the base \( b \) represents \( 1 - r \), where \( r \) is the rate of decrease.
- Here, \( b = 0.98 \), hence \( 1 - r = 0.98 \).
- Solving for \( r \), we get \( r = 1 - 0.98 = 0.02 \).
- Converting this to a percentage gives us a 2% decrease per month.
Therefore, the price of fuel A is decreasing by 2% per month.
### Part B: Analyzing the Price of Fuel B
The table below shows the price \( g(m) \) of fuel B after \( m \) months:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline m \text{ (number of months)} & 1 & 2 & 3 & 4 \\ \hline g(m) \text{ (price in dollars)} & 4.19 & 3.98 & 3.78 & 3.59 \\ \hline \end{array} \][/tex]
1. Calculate the percentage change in price for fuel B from month-to-month:
The percentage change from month \( m \) to month \( m+1 \) is calculated as:
[tex]\[ \text{Percentage change} = \left( \frac{g(m) - g(m+1)}{g(m)} \right) \times 100 \][/tex]
- From month 1 to month 2:
[tex]\[ \left( \frac{4.19 - 3.98}{4.19} \right) \times 100 = 5.0119\% \][/tex]
- From month 2 to month 3:
[tex]\[ \left( \frac{3.98 - 3.78}{3.98} \right) \times 100 = 5.0251\% \][/tex]
- From month 3 to month 4:
[tex]\[ \left( \frac{3.78 - 3.59}{3.78} \right) \times 100 = 5.0265\% \][/tex]
The percentage changes are approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \).
2. Find the greatest percentage change:
- Among the calculated percentage changes, the greatest percentage change is \( 5.0265\% \), which occurred from month 3 to month 4.
### Comparisons and Conclusion:
- Fuel A: The price is decreasing steadily by \( 2\% \) per month.
- Fuel B: The price has recorded percentage changes of approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \) over the previous months.
Conclusion: Fuel B has recorded a greater percentage change in price over the previous month compared to Fuel A. The maximum percentage change for Fuel B recorded is approximately 5.0265%, which is greater than the steady 2% decrease per month for Fuel A.
### Part A: Analyzing the Price of Fuel A
The price \( f(x) \) of fuel A after \( x \) months is given by the function:
[tex]\[ f(x) = 2.15(0.98)^x \][/tex]
1. Determine if the price is increasing or decreasing:
- The function \( f(x) = 2.15(0.98)^x \) is in the form of an exponential decay function because the base of the exponent (0.98) is less than 1.
- This indicates that the price of fuel A is decreasing over time.
2. Find the rate of percentage decrease per month:
- In exponential decay functions of the form \( f(x) = a \cdot b^x \), where \( 0 < b < 1 \), the base \( b \) represents \( 1 - r \), where \( r \) is the rate of decrease.
- Here, \( b = 0.98 \), hence \( 1 - r = 0.98 \).
- Solving for \( r \), we get \( r = 1 - 0.98 = 0.02 \).
- Converting this to a percentage gives us a 2% decrease per month.
Therefore, the price of fuel A is decreasing by 2% per month.
### Part B: Analyzing the Price of Fuel B
The table below shows the price \( g(m) \) of fuel B after \( m \) months:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline m \text{ (number of months)} & 1 & 2 & 3 & 4 \\ \hline g(m) \text{ (price in dollars)} & 4.19 & 3.98 & 3.78 & 3.59 \\ \hline \end{array} \][/tex]
1. Calculate the percentage change in price for fuel B from month-to-month:
The percentage change from month \( m \) to month \( m+1 \) is calculated as:
[tex]\[ \text{Percentage change} = \left( \frac{g(m) - g(m+1)}{g(m)} \right) \times 100 \][/tex]
- From month 1 to month 2:
[tex]\[ \left( \frac{4.19 - 3.98}{4.19} \right) \times 100 = 5.0119\% \][/tex]
- From month 2 to month 3:
[tex]\[ \left( \frac{3.98 - 3.78}{3.98} \right) \times 100 = 5.0251\% \][/tex]
- From month 3 to month 4:
[tex]\[ \left( \frac{3.78 - 3.59}{3.78} \right) \times 100 = 5.0265\% \][/tex]
The percentage changes are approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \).
2. Find the greatest percentage change:
- Among the calculated percentage changes, the greatest percentage change is \( 5.0265\% \), which occurred from month 3 to month 4.
### Comparisons and Conclusion:
- Fuel A: The price is decreasing steadily by \( 2\% \) per month.
- Fuel B: The price has recorded percentage changes of approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \) over the previous months.
Conclusion: Fuel B has recorded a greater percentage change in price over the previous month compared to Fuel A. The maximum percentage change for Fuel B recorded is approximately 5.0265%, which is greater than the steady 2% decrease per month for Fuel A.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.