Answer :
Alright, let's go through the problem step-by-step:
#### (a) Evaluate [tex]\(\frac{w(5) - w(0)}{5 - 0}\)[/tex] and [tex]\(\frac{w(15) - w(5)}{15 - 5}\)[/tex]
We have the table of values:
[tex]\[ \begin{array}{c|c|c|c|c|c|c|c} \hline t & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline Q & 1575 & 1660 & 1715 & 1750 & 1770 & 1775 & 1785 \\ \hline \end{array} \][/tex]
1. First, we calculate the average rate of change between [tex]\( t = 0 \)[/tex] and [tex]\( t = 5 \)[/tex]:
[tex]\[ \frac{w(5) - w(0)}{5 - 0} = \frac{1660 - 1575}{5 - 0} \][/tex]
[tex]\[ = \frac{85}{5} = 17.0 \][/tex]
2. Next, calculate the average rate of change between [tex]\( t = 5 \)[/tex] and [tex]\( t = 15 \)[/tex]:
[tex]\[ \frac{w(15) - w(5)}{15 - 5} = \frac{1750 - 1660}{15 - 5} \][/tex]
[tex]\[ = \frac{90}{10} = 9.0 \][/tex]
So, the average rates of change are:
[tex]\[ \frac{w(5) - w(0)}{5 - 0} = 17.0 \quad \text{ppb per year} \][/tex]
[tex]\[ \frac{w(15) - w(5)}{15 - 5} = 9.0 \quad \text{ppb per year} \][/tex]
#### (b) Compare your answers in (a). What does this tell you about atmospheric methane levels?
We notice that the rate of change [tex]\(\frac{w(5) - w(0)}{5 - 0}\)[/tex] is higher than the rate of change [tex]\(\frac{w(15) - w(5)}{15 - 5}\)[/tex].
This tells us that the methane level rises at a slower average rate from the years 5 to 15 than from the years 0 to 5.
To complete the explanation:
[tex]\[ \text{This tells us that the methane level rises at a slower average rate from the years} \, 0 \, \text{to} \, 5 \, \text{than from the years} \, 5 \, \text{to} \, 15. \][/tex]
So, the final answer is:
[tex]\[ \text{This tells us that the methane level rises at a slower average rate from the years} \, 0 \, \text{to} \, 5 \, \text{than from the years} \, 5 \, \text{to} \, 15. \][/tex]
#### (a) Evaluate [tex]\(\frac{w(5) - w(0)}{5 - 0}\)[/tex] and [tex]\(\frac{w(15) - w(5)}{15 - 5}\)[/tex]
We have the table of values:
[tex]\[ \begin{array}{c|c|c|c|c|c|c|c} \hline t & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline Q & 1575 & 1660 & 1715 & 1750 & 1770 & 1775 & 1785 \\ \hline \end{array} \][/tex]
1. First, we calculate the average rate of change between [tex]\( t = 0 \)[/tex] and [tex]\( t = 5 \)[/tex]:
[tex]\[ \frac{w(5) - w(0)}{5 - 0} = \frac{1660 - 1575}{5 - 0} \][/tex]
[tex]\[ = \frac{85}{5} = 17.0 \][/tex]
2. Next, calculate the average rate of change between [tex]\( t = 5 \)[/tex] and [tex]\( t = 15 \)[/tex]:
[tex]\[ \frac{w(15) - w(5)}{15 - 5} = \frac{1750 - 1660}{15 - 5} \][/tex]
[tex]\[ = \frac{90}{10} = 9.0 \][/tex]
So, the average rates of change are:
[tex]\[ \frac{w(5) - w(0)}{5 - 0} = 17.0 \quad \text{ppb per year} \][/tex]
[tex]\[ \frac{w(15) - w(5)}{15 - 5} = 9.0 \quad \text{ppb per year} \][/tex]
#### (b) Compare your answers in (a). What does this tell you about atmospheric methane levels?
We notice that the rate of change [tex]\(\frac{w(5) - w(0)}{5 - 0}\)[/tex] is higher than the rate of change [tex]\(\frac{w(15) - w(5)}{15 - 5}\)[/tex].
This tells us that the methane level rises at a slower average rate from the years 5 to 15 than from the years 0 to 5.
To complete the explanation:
[tex]\[ \text{This tells us that the methane level rises at a slower average rate from the years} \, 0 \, \text{to} \, 5 \, \text{than from the years} \, 5 \, \text{to} \, 15. \][/tex]
So, the final answer is:
[tex]\[ \text{This tells us that the methane level rises at a slower average rate from the years} \, 0 \, \text{to} \, 5 \, \text{than from the years} \, 5 \, \text{to} \, 15. \][/tex]
