Answer :
To solve this problem, let's carefully evaluate both Pierre's and Scott's equations based on the given conditions.
We need to write an exponential growth equation that has:
- A starting value of 300
- A growth rate of 2%
### Evaluation of Pierre's Equation:
Pierre suggests the equation:
[tex]\[ y = 300(1.02)^x \][/tex]
Let's break this down:
1. The initial value is 300, which matches the problem statement.
2. The growth rate is 2%. To convert a percentage to a decimal, we divide by 100:
[tex]\[ 2\% = \frac{2}{100} = 0.02 \][/tex]
3. An exponential growth formula typically looks like:
[tex]\[ y = \text{initial value} \times (1 + \text{growth rate as a decimal})^x \][/tex]
4. Therefore, the growth rate as a base should be:
[tex]\[ 1 + 0.02 = 1.02 \][/tex]
Pierre's equation matches this form perfectly:
[tex]\[ y = 300(1.02)^x \][/tex]
Hence, Pierre's equation correctly represents the given conditions, and Pierre is right.
### Evaluation of Scott's Equation:
Scott suggests the equation:
[tex]\[ y = 300(1.2)^x \][/tex]
Let's analyze this:
1. The initial value is 300, which again matches the problem statement.
2. According to Scott, the base of the exponent is 1.2.
3. However, the correct base should be:
[tex]\[ 1 + \text{growth rate as a decimal} = 1 + 0.02 = 1.02 \][/tex]
Scott's base of 1.2 does not match the 1.02 required for a 2% growth rate. Therefore, Scott's equation does not accurately represent the given conditions, and Scott is incorrect.
### Conclusion:
- Pierre is right. His equation correctly accounts for a 2% growth rate by using the base 1.02.
- Scott is incorrect because his base of 1.2 does not represent a 2% growth rate.
So, the correct evaluation is as follows:
a. Pierre is right. \( 2\% \) written as a decimal is \( 0.02 \), so the base should be \( 1 + 0.02 = 1.02 \). Hence, Pierre's equation:
[tex]\[ y = 300(1.02)^x \][/tex]
correctly represents the problem.
Scott's equation does not reflect the correct growth rate and thus is:
[tex]\[ y = 300(1.2)^x \][/tex]
incorrect for a 2% growth rate.
We need to write an exponential growth equation that has:
- A starting value of 300
- A growth rate of 2%
### Evaluation of Pierre's Equation:
Pierre suggests the equation:
[tex]\[ y = 300(1.02)^x \][/tex]
Let's break this down:
1. The initial value is 300, which matches the problem statement.
2. The growth rate is 2%. To convert a percentage to a decimal, we divide by 100:
[tex]\[ 2\% = \frac{2}{100} = 0.02 \][/tex]
3. An exponential growth formula typically looks like:
[tex]\[ y = \text{initial value} \times (1 + \text{growth rate as a decimal})^x \][/tex]
4. Therefore, the growth rate as a base should be:
[tex]\[ 1 + 0.02 = 1.02 \][/tex]
Pierre's equation matches this form perfectly:
[tex]\[ y = 300(1.02)^x \][/tex]
Hence, Pierre's equation correctly represents the given conditions, and Pierre is right.
### Evaluation of Scott's Equation:
Scott suggests the equation:
[tex]\[ y = 300(1.2)^x \][/tex]
Let's analyze this:
1. The initial value is 300, which again matches the problem statement.
2. According to Scott, the base of the exponent is 1.2.
3. However, the correct base should be:
[tex]\[ 1 + \text{growth rate as a decimal} = 1 + 0.02 = 1.02 \][/tex]
Scott's base of 1.2 does not match the 1.02 required for a 2% growth rate. Therefore, Scott's equation does not accurately represent the given conditions, and Scott is incorrect.
### Conclusion:
- Pierre is right. His equation correctly accounts for a 2% growth rate by using the base 1.02.
- Scott is incorrect because his base of 1.2 does not represent a 2% growth rate.
So, the correct evaluation is as follows:
a. Pierre is right. \( 2\% \) written as a decimal is \( 0.02 \), so the base should be \( 1 + 0.02 = 1.02 \). Hence, Pierre's equation:
[tex]\[ y = 300(1.02)^x \][/tex]
correctly represents the problem.
Scott's equation does not reflect the correct growth rate and thus is:
[tex]\[ y = 300(1.2)^x \][/tex]
incorrect for a 2% growth rate.
